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At every step of Insertion algorithm, to find the proper location of the current element(a[j]) in already sorted subarray whenever we are shifting right a[i] (as a[j] is smaller than probed element a[i]) we are deleting the inversion pair (a[i],a[j]). so If there are at most n inversion pair insertion sort needs atmost n steps to correct all the inversion pair hence the time complexity is O(n).
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Inversions here means we gotta swap the position of elements Ai and Aj if Ai>Aj and i
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Go to this link, it is well explained there : https://gateoverflow.in/43576/gate2003-62