@dipit ..
Yup you are correct..
Say, no of rows = M and no. of cols = N,
Time complexity = sum over all i (1 to M} { N*(M+N-i) }
= M * N * (M + 2N - 1) /2
On Jan 11, 2:19 pm, Dipit Grover <[email protected]> wrote:
> @Lucifer : I came up with a similar algorithm as yours but I dont
> understand your complexity analysis : " sum over all i (1 to M} { i*(M+N-i)}
> " .
>
> Shouldnt it be " M * sum over all i(1 to N) {(M+N-i)} " ? M= no of
> columns, N= no of rows . Since we always have the min element at the 0th
> column of the next row for each element of the current row.
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