@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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