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Merge initial order

hello friends today we are going to see how to merge two sorted arrays into a third sorted array so let's see see this is the array one and the array two these two arrays are sorted and we want to merge these two sorted arrays into this result array which is also sorted so obviously one option is to merge these arrays and then sort the resultant array but that will take of n log n time complexity but we have a solution which works in of n time complexity so let's see what is that solution in this solution we make use of the salted nature of this array 1 and array 2 - lets see the rules so this is the pseudocode for merging the sorted arrays in the pseudocode variable i will trace the array 1 and variable j will trace the array 2 and you have to declare the resultant array of the size s 1 plus s 2 means s 1 is the size of array 1 and s 2 is the size of array 2 so the resultant array will be of size s 1 plus s 2 ok now let's see what is the rule see while I is less than s 1 and J is less than s 2 means I will go up to s 1 the size of array 1 and J will go up to s 2 the size of array 2 now inside while if array 1 of I is less than a rating of J that means you have to find out at this instant at this position which element is of less value means whether array 1 of I is less than J 2 of J that's what you are checking and the element which is less will go to the final area that is the resultant array so see here if array 1 of I is less than array 2 of J then result of k equal to array 1 of I means in the resultant array the element at I will be shifted means in this case because 1 is less than 2 1 will go to the final array okay and after it goes to the final array see here you have to do I plus plus means the element which shifts to the resultant array is the element pointed by I and that is why I will increment now if the element at J will get shifted to the resultant array then J would have incremented means once we shift that element you have to increment that respect you variable okay see now because every one of I is shifted to the resultant array i will increment that is i plus plus so i will be incremented and obviously you have to increment K because now at position K element 1 is stored now it will increment to the next position I will do index here 0 1 2 3 and 11 okay so K is incremented to the next position to hold the next element now let's solve the problem with this procedure now see check whether array 1 of I is less than added to of J 3 is less than 2 no so this condition is not satisfied we will go in ends that means element and J is lesser okay means because array 1 of I is not less than of je we go in else and inside else see era two of J will be shifted in the resultant array okay so what is era two of J that is to the simple rule is that the lesser element will be shifted to the resultant adding okay so to each shifted to the resultant array and as per the rule the element at J is shifted to the resultant array so now jaylee increment and then k will be incremented to hold the next element now let's solve it first see check whether 3 is less than 4 yes it is so 3 will be shifted to the resurrected array and I will be incremented okay now whether 5 is less than 4 no but 4 is less than 5 so 4 will be shifted to the resulted array and J is incremented now whether 5 is less than 8 yes it is so file will be shifted to the resultant array and I is incremented see don't forget to increment K I just forgot to increment it from this position you just keep that in mind every time you have to increment key okay so whether 7 is less than it yes it is so 7 it shifted to the resultant array and I is incremented and again K is also implemented okay now whether 9 is less than it no but 8 is less than 9 so we go in else that means 8 will be shifted to the resultant array and J is incremented okay now check whether 9 is less than 10 yes it is so 9 is shifted to the resultant array and I is incremented so I becomes 5 and when I becomes 5 see here I is less than s 1 condition fails ok because s 1 is fine the size of the first array is 5 and 5 is not less than 5 ok now I is 5 and s 1 is also 5 so 5 is not less than 5 so this condition fails and as this is the and condition we come out of this while loop ok we come out of the while loop and obviously you have to copy paste all the remaining elements from the second array C because I is exhausted now means the first array is over now obviously you have to copy paste all the elements remaining in the second array as it is in the final array ok so all the elements from J to the last position is called are copy pasted to the final array so 10 11 12 and 16 are copy pasted so that is written here see while I is less than s 1 but this will not work because I is not less than s 1 I is equal to s 1 so this while will fail and we go to the next while loop while J is less than s 2 while J is less than s 2 you have to shift all the elements from array 2 to the resultant array okay and K will be 2 the last index so all the elements are copy pasted to the final array now in some case J will exhaust first if J is exhausted first means J reaches the end of the array and I is still somewhere in the middle of the array then all the elements remaining in the first Arabic of the vested in that case ok so whatever the array elements are remaining are copy pasted into the final array now as you can see here the time complexity will be of s1 plus s2 c of n means n equal to s1 plus s2 means the size of the resultant array because we have traversed the resultant array size right and we can extend this algorithm to merge multiple sorted arrays multiple sorted arrays into ones moderator hey friends please subscribe to my channel as I post algorithm videos every day and if you want a video on any particular topic then please mention in the comment below thank you