dear students like And subscribe my channel this is mer sort algorithm mer sort algorithm using DDE and Conquer technique this examples you can solve using divid and Conquer technique in m s first this array split into two subgroups your size of R is n equal to n = 7 me this array split into two subgroups me that is n by 2 N by2 equal to either three or 4 but this is given array 70 20 30 40 10 50 then 60 this is array array is splitted into two sub groups that is n by 2 means one side there are four elements another side three elements and this m s you can solve using top down approach M top to bottom size of AR is seven and the Seven is splitted into two subgroups me there are one side four elements and other side three elements 70 20 30 40 10 50 then 60 miss this low value and this is high value low value and high value for left side part low value is same right side part high value is same left side part low value to Mid and right side part mid + 1 to high again this string split into two sub me left part and right part split into two subgroups that is n by 4 and in left part there are three elements me split this string to left side part one side two elements another side one element left part as well as right part and this string we to split up to single digit if single digit is there then we cannot split 10 50 60 again within this subgroup there are two elements 2 2 2 then one me one means we can keep as it is this is two array two elements then split into two subgroups again within this two elements then split into two subgroups 60 only one element is there we can keep as it is this is 70 20 30 40 if numbers are there me we can arrange in ascending or descending order but by default we can arrange in as ascending order now combine these two and write in ascending order now this is 20 70 left part sub solution similarly right part subsolutions again arrange in ascending order me this is 30 40 similarly for left part subgroups this is 10 50 next solution of left part solution of left part arrange in ascending order four elements this is 20 30 40 70 and solution of right side part solution of right side part this two elements and one more elements arrange this in ascending order me 10 50 60 then M left part and right part and arrange in ascending order total Els 7 ARR in ascending order this is 10 20 30 40 50 60 then this is 770 this is called as marot n is splitted into two subgroups again left for splited into two subgroups again subgroup split into two subgroups if single element is there then we can merge and arrange in ascending order left solution right solution left part solution similarly for right side potion this is called as more sort and this elements digits are there digits are there me we can arrange in ascending or descending order similarly we can arrange word in alphabetical order similarly we can arrange here this is mar sort algorithm for the word question for this word all letters we to arrange either in Al we have to arrange in alphabetical order right alphabetical order means from a to zed and if I want to arrange A to Z again this array splited into two subgroups here size of array is size of array is eight if size of array is eight then this spring and this string splitted into two two group two sub groups N by2 N by2 means four four elements left part q u e s t i o n left part for this four elements and right side part for this four elements me this n is splitted into two subgroups four on this side also four plus q u p yes right side part t i o n again left part there are four elements if there are four elements again four elements split into two subgroups that is split into two subgroups means two two elements q u then e s right side part same two two elements so the for is four here t i o n again left subgroup and right subgroup there are two two elements me split into single group s single element left part and similarly left part with the left part right side me this is q u e s same process apply for right side part are to split this array up to single digit if single digit is there me you can stop this is t i o again AR is splitted now have to merge m solution of left subgroups and then merge in alphabetical arrange in alphabetical order these two elements means first is Q then U similarly within left side right part this is e s same process you can apply for right side part this is i t then n o solution of left part and solution of right part solution of left part as well as solution of right part solution of left part arrange in ascending order arranging ascending order means e q yes then u e q s u then they said i n o t then solution of origin problem mge left part as well as right part here there are size of arrays four and solution of left part as well as right part then arranging alphabetical order Miss e i n o q s t u this is in alphabetical order this is called as M sort apply M sort algorithm for the word to arrange in alphabetical order next dear students you can see the time complexity of time complexity of MOT and for the examination purpose time complexity is important in all cases time complexity is same in time complexity time complexity time complexity of mord time complexity of m t of n equal to when we can split array if size of array is more than 1 me if n = to 1 then time complexity is zero otherwise otherwise this is T of n by 2 + t n by 2 + n me n is splitted into two subgroups n by2 n by2 again similarly this n by2 left part as well as right part splited into two subgroups me so to solve this expression this T of n equal to T of n by2 + T of n by2 + y for this you can write 2 T of n by 2 + n right this is if n n is splitted into two subgroups again this subgroup splitted into two subgroups Miss T of n by2 = to 2 t n by n by 2 by 2 + n by 2 me for this you can write 2 t n by 2 ^ 2 + n by 2 and put this value in equation one put value of T of ny2 put value of T of n by2 in equation one this within this equation we have to put this value this two as it is this two t n by 2^ 2 + n by 2 + n as it is 2 into 2 you can write 2 ^ 2 t n by 2 ^ 2 + 2 N by 2 + n this 2 two cancel this is 2 n 2 the^ 2 T of n y 2 ^ 2 + 2 N again this is n by2 again n by2 splited into two subgroups similarly n by2 splited into two subgroups and N by2 splitted into two subgroups me direct we can write m t of n by 2^ 2 equal to 2 T 2 T this n is split into two so n by 2 N by 2 by 2 Miss n by 2 to the power 3 + n by 2 the^ 2 again put this value in equation 2 was T of n by 2 ^ 2 this value put in equation 2 but this is 2 ^ 2 constant keep as it is 2^ 2 t n by 2^ 3 + n by 2 ^ 2 + 2 N again we can multiply this 2 ^ 2 2 ^ 3 t n by 2^ 3 + 2 ^ 2 N by 2 ^ 2 plus this 2 n 2^ 2 2 ^ 2 is cancel for this you can write T of n = 2 power 3 n by 2^ 3 + 3 n right again we can solve again 2^ n by 3 2^ n by 3 we can split into two subgroups up to n we can split this is up to T of n = to 2 already n is there Me 2 the Power i t of n by 2 power I plus I N up to I time and 2 the^ I equal to what is 2^ I 2^ i = n and uh time complexity if n = 1 me time complexity is zero T of n = 2^ i t of n by n + i n t of n by is 1 and T of n is zero me T of time complexity i n what is I to find value of I value of I here 2^ I = to Y you can solve this equation using taking log on both side taking log on both side means log 2^ I = log of n here base nothing is there means by default Bas is two Bas is two and two is means this is called as one log base 2 2 = 1 using this formula this is I = log of N and this is equation three put this value of I value of I in equation 3 in equation 3 me here time complexity T of n equal to for this we can write n log of n must time complexity you can represent it big of Y log of Y in all three cases time complexity is same big of n Theta of N and Omega of n this is called as Mar dear students all of you I think understood all of you subscribe and share this Channel with your friends