let x by the number that is more repeated after at most k operations , let Fx by the frecuency of x after at most k operations (Fx is the solution to the problem) let A by the array sorted let S by an array where S[i] = A [0] + A [1] +. . . + A[i] is easy to see that x is an element of A whitout make operations, if A[i] = x, then Fx = (i — j + 1) where j is the minimum index where it holds that ( (i — j + 1) * A [i] ) — (S [i] — S [j-1]) <= k, to find j, a binary search is done from [0 .... i].