What doubt you have in Insertion Sort?

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+1 vote

Here "j" loop will run "n-1" times and for "while" loop suppose you reach upto j=n and A[n] = smallest of all the value array so we have to go checcking from A[N-1] to A[1] therefore in worst case inner loop will run "n-1" time.

so in worst case time complexity will be O( (n-1)(n-1) ) = O(n^{2}).

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