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hi my friend welcome to joystick first of all I would like to thank you for all the love that you have showered on my channel and back with this video in which we are going to solve using dynamic programming that takes justification problem now while you are watching this video you are going through a serious transformation that means you are levelling up in your dynamic programming skills because text justification problem is one of the most complex dynamic programming problems and I have tried my best to explain it to you in a very easy way so I urge you to stay till the end I urge you to keep watching this video till the end and it is going to make sense to you and I'm I'm I'm sure that you are going to move one level up in your dynamic programming skills so let's move to the problem first before moving on to the problem I urge you again to subscribe to my channel and hit the bell icon so that you don't miss out on any of my videos of dynamic programming now let's move to the problem so that takes the justification problem is more mostly witnessed in Microsoft Word so while you are typing in Microsoft Word and you have the in margins set and you must have noticed too that the microsoft word automatically adjusts your line ending and once you type the Pera the entire peril looks very neat and very properly arranged the words are very much properly arranged in each and every line so ms word makes use of that takes two justification algorithm which is what we are going to create here using dynamic programming so it is all about putting line breaks at the correct places such that the cost is minimum the cost of the remaining spaces I am talking about you are given a sequence of words in this problem and also the limit of characters that you can put in one single line so to solve the text a justification problem all you need sequence of words and the limit of characters that you can put in one single line these two are the pillars okay of the texture justification problem you need to create an algorithm which places the word in each line neatly such that in looks justifiable and proper like I already said Microsoft Word does that for you so it's it's a clear example of using the text to justification algorithm let's proceed now in this problem what we are going to do we have the sequence of words so word wrap problem solved by joystick is the sequence of words that we are going to arrange okay in lines and each line has a maximum length of 15 characters that means the number of characters that are allowed in every single line is 15 not more than 15 we can place in one single line so that is one of the conditions like I already talked about that we have and based on this sequence of words and this maximum length per line we will solve our type text a justification problem attention so before I proceed further I need you to sincerely focus on this line so we have an arrangement let us say we have got five words five six times a 3 times B 4 5 times C double D and full-time C these are just random words consider this as a sequence of words that is for the sake of your understanding of the cost and the maximum characters per line is 10 okay hypothetical scenario okay the normal approach that you are going to follow is that you are going to stuff as many words as possible in one line so this is going to be the example of normal approach in which you will have the word starting with a the word starting with B in one line the word starting with C and the word starting with D in second line and finally the final line of yours will contain the words starting with E so this is going to be the normal approach as you can see that you are stuffing as many words as possible in one single line okay the optimized approach is going to come now but before that we are going to calculate the cost so the cost is going to calculate to be calculated by summing up the number of remaining spaces on each line let's see what is that so zero is the number of spaces that the number of remaining spaces from line 1 2 is going to be the number of remaining spaces from line 2 and 6 is going to be the number of remaining spaces from line 3 as you can see the maximum length allowed on each line is 10 if you check the first line we have the number of characters as 6 times a and 3 times P so 6 plus 3 is 9 plus 1 space between that two words so that is going to come out as 10 10 minus 10 is zero so the number of remaining spaces is 0 in the second line it's 5 times C plus 2 times B so 5 plus 7 now 5 plus 2 is 7 with one space between the two words so it that makes it 8 10 minus 8 is 2 similarly 6 is what we get from the third line now let's move to the optimized approach so the optimized approach will have the word starting with a in line 1 the word starting with B and the word starting with C in line - the word starting with T and the word starting with E in line 3 this is the optimized approach that Microsoft Word is going to follow and that takes two justification algorithm is also going to give you this answer when the sequence of words is this and the maximum characters per line is 10 let's analyze the cost of this approach as well so 4 is the number of remaining spaces that I get from line 1 1 is the number of remaining spaces that I get from line to and number and three is the number of remaining spaces that I get from line three so as you can see from the first line itself we have six times a so six is the number of characters ten six subtracted from ten you get four so likewise we have one entry but you are seeing that the total cost which is coming out in both these approaches is 8 which is the same so how the second one is going to be an optimized approach how we are going to discriminate between these two between the cost is C so here the square of the number comes into picture we are going to measure the cost by summing of the squire's of the numbers it's a not strict it's not a strict mandate or not a rule that you have to use the square of a number you can make use of cube as well but if you are going to make the use of only numbers then the cost is going to come out as same which is not going to solve your purpose which is going to confuse this algorithm so let's move to the second slide and see how the squares of the numbers solve our cost problem so cost in terms of square we have the same arrangement then here's the normal approach like we discussed in the previous slide the cost is going to come out as 40 so zero was the number of remaining spaces from line 1 2 was the number of remaining spaces from line 2 so the square of 2 is 4 and 6 is the number of remaining spaces from line 3 the square of 6 is 36 the sum of the squares of these numbers is going to come out as 40 now let's look into the cost which is calculated by following the optimized approach so here is our optimized approach we have the arrangement in front of us and let's look at the cost so 4 was the number of remaining spaces nine-one-one was the number of remaining spaces from line two and three was the number of remaining spaces from line three so for square 16 square of 1 is 1 and square of 3 is 9 if we sum all of them up then we get 26 and 26 is the minimum of the two approaches the costs that have been calculated in two approaches and based on this number 26 we can say that yes the second one is the optimized approach and this is what we are going to use in our solution so now let's move to solving the problem solving the takes the justification problem so here we are at our set up this is our cost matrix which we are going to fill with values which we are going to calculate by by analysis by thorough analysis and there is one table here in the right this table is just here for the calculation purposes it is just here for the sake of your understanding and it is going to store some intermediate calculations this is not necessary when you are solving it at home all by yourself you can just use your mind to make it happen but just for the sake of your understanding I have created this table okay we are going to delete it after it is done let's go through the parts of this table so we have the index over here and we have the word in the second column so you can see that for each of the indexes I have listed the word over here and also I have mentioned their respective lengths in front of them then we have these two column spaces and remaining spaces which we are going to use when we are going to calculate the values of this matrix so you are going to witnesses so let's start by let's start from this cell let's start by filling this matrix we are going to start from this cell so as you can see the rows of this matrix are also the words given to us and the columns of this matrix are also the words given to us okay so we are going to start from this cell so 0 comma 0 that means it means that we have the word at index 0 in first life that means only the word at index 0 we have to place at line 1 and let's calculate the cost of it so how we are going to calculate the cost as you know that it is all about remaining spaces so what we are going to do we are going to subtract this length from the maximum length allowed so for subtracted from 15 is going to give us level so I'm going to put 11 over here and since we have always discussed to that we need to calculate the square of the remaining spaces so I am going to place one 21 here which is the square of number 11 so when we have only word in the first line the cost is going to come out as 121 now let's move to the set so 0 to 1 that means word and wrap in one single line now the length is going to come out as 8 and there is going to be one space between them so it plus 1 9 9 subtracted from 15 is going to give us the remaining spaces as 6 now the square of 6 is what we are going to place here so 36 is the value of this cell now let's move to this cell so 0 2 2 that means word 2 problem now let's sum that sum them up so 4 plus 4 is 8 plus 7 is 15 now we are going to have two spaces also so one space I am counting between word and rap and another space and counting between rap and problem so it's going to give me the total length of characters as 70 which breaches the maximum length so it is not allowed so when such kind of scenario rises in this problem during solving this problem all we are going to do here is place infinity so we are going to place the value as infinity over here that means this denotes that this is not possible this is a gone case this is this is out of our hands maybe in a different universe so it's possible but not in ours okay now we move to this cell from 0 to 3 that means from here to here were drag problems solved clearly you can see the combined length is going to come as 21 which breaches the maximum length so again we are going to fill this cell with infinity we move to this cell now from 0 to fourth word wrap problem solved by again the combined length is going to come out as 23 which breaches 15 so the value is going to be infinity similarly the value of this cell also is going to be infinity so we come to the next row now we are going to start from this cell because we are talking about the word at index 1 which is a wrap which cannot be before this word at index 0 which itself is word so we are going to start from cell 1 so 1 2 1 that means wrap we are going to calculate the remaining spaces to avoid confusion I am going to omit these 2 values over here and we will be calculating a new value so let's omit this value as well 15 minus 4 is going to come out as again as 11 so the value here will be 121 which is the squire of left moving on to the cell 1 to 2 that means wrap to problems so the combined the length is going to be 11 plus 1 space between wrapping problem so it's going to be 12 15 minus 12 is going to give you 3 and hence the value here is going to appear as 9 the square of 3 moving on to this cell 1 2 3 as you can see the combined the value of this is going to be greater than 15 4 plus 7 is 11 plus 6 is 17 so this case is not possible hence we are going to populate infinity over here similarly now in this cell from 1 to 4 as you can see again the value is going to the sum of the lengths of these word words is going to cross 15 so this combination will not be possible on one line hence we are again going to place infinity over here similarly in this cell also we are going to place infinity moving on to the next row we are going to start from here because now we are starting from this word which is problem and problem appears afterward and drag so the values in these two cells cannot be calculated when we are starting from problem it is obvious so now here 2 2 2 the length is 7 let me delete these values to avoid confusion 15 minus 7 is 8 and hence I am going to get the value here as 64 moving on to the next cell from 2 to 3 we are going to get the value as 13 and with 1 space it it is going to become 14 and the remaining spaces is going to be 1 so square of 1 is 1 hence I'm going to place one over here now from 2 to 4 we have to calculate from 2 to 4 as you can see the combined lengths the combined length is going to be 15 and 2 spaces 1 between a problem and solved and another one between salt and buy so 15 plus 2 it's going to be 17 it reaches 15 hence it cannot be calculated and we are going to populate the sell faith infinity moving on to the next row which starts from index 3 the world at index 3 sold we will be starting from here so 3 comma 3 is going to make it 6 to avoid confusion I am going to delete these cells so 15 minus 6 is going to be 9 and I will place the value 81 over here now 3 2 4 is going to be our next cell calculation solved by 6 plus 2 is 8 plus 1 space between them so 15 minus 8 is 7 so I'm going to place 7 over here I'm sorry it is going to be 9 6 plus 2 is 8 plus 1 9 so 15 minus 9 is going to be six not seven and I will place 36 over here now for this cell it will be 3 to 5 so from 3 to 5 the combined length is going to come as 18 plus 2 spaces so it's going to be 28 breaches 15 hence we are going to populate this cell with infinity as well let me or my T cells now we are going to move to row which starts with index four we will be starting from this sin the word at index 4 is by as you can see the length is - 15 - 2 is 13 13 is the number of remaining spaces and we will place 169 which is the square of 13 over here moving to this sin this cell will have will cover the words from 4 to 5 from 4 to 5 that means by and Joystiq will be covered in in this way so 10 plus 2 is 12 plus 1 space between them so it's going to be 13 15 minus 13 is 2 so the value which we are going to get here will be 4 now we will move to the final row which starts with index 5 over here so if as you can see the length is 10 so when you are going to subtract 10 from 15 you are going to get the length as 5 and the value which will be populated here is the square of 5 which is 25 so this completes our cost of matrix the cost to matrix is complete but this is not the final solution this is the information that we are going to use to calculate the final solution that means the cost and arrangement let's see how we do that so here we are with our two arrays one of which represents the cost so this array over here is going to carry the cost the minimum cost for the arrangement to be neat and precise and proper and over here this is our arrangement array so this is going to store the the arrangement positions for the words to be neat and proper so we are going to use this array to place the words neatly just like microsoft word does and we are going to use this array to calculate the minimum cost let me please this in position we put four and five over here and here we have our cost matrix which we are going to use all you need to do is focus up a little because the problem the solution from here is going to be a little tricky but I have tried my best to explain it to you in an easy way so here are our two pointers I'm going to place I over here and what do these mean you are going to understand once we move ahead so I actually represents the row and J represents the column from this cost matrix so we are going to start from right and move towards the left so let's start solving let's start filling this cost array in fact so when I is five and J is five let's check the value over here so I is five and J is five the value is 25 we are going to fill 25 as the value here now this is the minimum cost of a sub-problem now the sub problem states that it here it only has the word chewy stick okay forget about these words because when we are solving a problem using dynamic programming we have to talk about subproblems so this is a sub problem so what the minimum cost is going to be then we only have the word chewy stick with us and the maximum limit on the line is 15 it's obviously going to be 25 because 15 minus 10 10 is the length of joystick so 15 minus 10 is going to be 5 square of 5 is going to be 25 what we are going to store here is that if this word is going to be in one line or till that shishun how many words are going to be in one line and how many words are going to be in the next line that information we are going to store here so Joystiq is obviously going to be in one line so I am going to put six over here now this means that number five the word ID index five is going to be on one line excluding the word add index six now in our problem there is no word at index six okay so that is why I have kept it as a as a stopper okay now we move for the left so I move I to the left now I is 4 and J is 5 so let's see what is the value I is 4 and J is 5 so the value over here is 4 I'm going to place pore over here okay and the value over here is going to be 6 that means now our sub-problem has by and Joystiq as the words as you can see the minimum cost is going to come as for which we have taken from here and they can be placed in one line only and that is why I have kept a 6 here this means that 4 and 5 so the words at index 4 & 5 can be placed in one line with 6 as the stopper so this indicates like this like this particular cell it indicates that the minimum cost is going to be 4 and the index at the word at index 4 and the word at index 5 which are by in Joystiq are going to be in one line so think it that way visualize the subproblems it will be easy for you it will start making sense automatically now let's let's move I to the left okay so now I have I as 3 and J is 5 so 3 and 5 as you can it is infinity so when we have stumbled into infinity that means the solution is calling for a split okay what does that mean it means that three to five that means solved by joystick and not people in one line that is why we have encountered infinity over here so this is calling for a split this is where the problem becomes tricky so understand now okay watch it carefully so now the split can happen either at five or four that means either joystick can move to next line and solved by are going to remain in the first line or by and Joystiq are going to move to next line and sold is going to remain in the first line we are going to determine that based on the cost so when the split is at five that means J equal to five so the cost is going to be so what is the cost at J equal to five it is 25 okay and then so the cost of three to four because the split is at 5 so 3 to 4 that means 3 to 4 the cost is going to come out as 36 the total cost will be 60 so whenever the split is at 5 that means Joystiq went to east ik is moving to the next line then the cost is going to come out as 61 now let's calculate the cost when the split occurs at PI this means PI and Joystiq are in the second line and sold is in the first line so let's move I over here so the value is going to come out as four over here and what will be the value at 3 comma 3 so the value at 3 comma 3 that means I is 3 JS 3 the value is coming out as 81 so the total value is going to become as 85 85 is greater than 61 so we will have to accept the splitting at 5 and the value that we are going to put here will be 61 here we are going to put the value as 5 this means that words at index 3 and 4 are in one line and from 5 onwards we will have to go back in the array to find whatever the word will be there after okay J goes back to its position which is over here all right and I moves for the left I hope you have caught it if you have not got it then comment below I'll answer your comments I'll answer your questions within the next 24 hours all right so when I is at 2 let's see what value 2 and 5 give so 2 to 5 it's infinity that means the solution is calling for a split so let's plate at 5 what the value at 2 and 4 is going to be that means 2 & 4 it's going to be a key in infinity that means from problem - bye ok we cannot put them in one line so split at 5 is not at all possible that means we'll have to move J one step left over here now let's check what the value at 2 & 4 is so 2 & 4 it's again infinity so this again calls for a split so split can happen either at 4 or 3 let's see of what will be the value when the split happens at 4 so when J is going to be 4 okay the value here is 4 plus the value from 2 to 3 so the value at 2 to 3 is 1 it is going to come out as 5 now I'm going to move J further left that means I am checking for the value when the split happens at 3 so when J is equal to 3 that means 61 Plus what will be the value at 2 and 2 it is going to be 64 so the combined value is going to come out as 125 now nove 125 is greater than 5 hence we will have to settle with the split at 4 so I'm going to place four over here and the value which I am going to place over here will be 5 all right amazing so J goes back to its base position I moves one cell left ok so now we have 1 2 5 let's see what is the value at 1 2 5 it's infinity so that split is not possible because the value at 1 2 4 is also infinity therefore we will move J one step left over here now 1/2 for the split AK 4 is also not possible because 1 2 3 the values infinity that means from rap to salt we cannot place in one line so J moves one step left so 1 2 3 ok it's also infinity but the split is possible either at J equals 2/3 or J equals to 2 okay because we have the value of 1/2 to calculated that means we can put happen problem in one line all right so what happens when J split happens at J equals 2/3 okay the value will be calculated as 61 because at J equal to 3 the value is 61 ok and one to two we have to check so one to two the value is going to be nine all right so the sum is going to come out come as 70 now we move J left to check what will be the value of the split happens like - okay so this is going to be five plus what is the value at one two one oh my god it is 121 so the value is going to be 126 now as you can see 126 is bigger than 70 so we will have to settle with split at T equals 2/3 so I'm going to put the value here as 70 the split that means the number of words till the split are going to be one in two so I'm going to put three over here I hope you are following this if not then put your doubts in the comment section I'll answer them within 24 hours J goes back to it's a base position and I moves one step left now check this up we are at our final step now now zero to five now zero to five cannot be possible because it's infinity cannot be placed in one line okay zero to four is also not possible zero to three is also not possible because the split cannot happen because zero to two is a keen infinity only they only sell only position at which the split is possible is at J equals to two so I am going to bring two I'm going to bring J directly over here okay so I GE equals two to the split is going to be possible because 0 & 1 can be placed in one single line so J equals to 2 let's calculate the value the cell value is already 5 and 0 2 1 let's see what the value is going to be it's going to be 36 all right so the value is going to come out as 41 the sum is going to come out as 41 now we move J one cell left over here so J equals to 1 okay we have 70 as the value and what is the value at 0 and 0 it is 121 all right so let me add 70 to 120 1 it becomes 191 now 191 is greater than 41 41 is minimum that means the correct split that is going to happen is going to happen at J equals to 2 and that is why we are going to put 2 here in our arrangement array and the minimum cost to that we are going to put is 41 now 41 is our answer that means when we whenever we have the following sequence of words were wrap problems solved by joystick and the maximum characters allowed in one line is 15 then the minimum cost is going to come out as 41 now let's use this arrangement array to place these words in the line and then you are going to fitness weather and you are going to see it yourself that the arrangement is going to be very neat and proper as it is done by Microsoft Word okay so from here we are going to start this the value here is 2 this means that 0 and 1 are going to be one line so word and rap are going to be in one line so let me put them here holdin rap okay so here the pointer states that we have to move to sell to alright so we are going to use a shade we move to two now here what it is same value in this cell is four that means words at index two and three are going to be in line two so words at two and three are problem solved so problem and salt are going to be in line number two I'm going to mark this as our path and since these two words are going to be in line two hands decorate them let's move to fourth alright so here what does this state this means that words at index 4 & 5 are going to be in third line the words at four and five are pi and Joystiq pi and choice take are going to be in the third line and this is our arrangement this is our neat and clean and proper arrangement using the text justification algorithm created using dynamic programming now let's go and watch the algorithm so this is the algorithm over here FC FC is our cost array so it's going to be minimum of our matrix I J minus 1 plus our cost array the value at index J where J is equal to I plus 1 in the length the time complexity of this algorithm is going to be zero fit in braces and Squire so I hope you enjoyed watching this video remember move one level up in algorithms through joystick I'll see you in the next video of Joey steak thank you very much for watching this video ciao ciao