Editorial for Back From Summer '19 P1: Winnie's Orphans

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Author: Ninjaclasher

We can loop through each orphanage, keeping a running minimum of which orphanage has the least number of children with a cuteness value of $1$ $1$ or $10$ $10$ . If the current orphanage has fewer such children than the running minimum, we update the best solution to the current orphanage's index. Otherwise, we move to the next orphanage.

Time Complexity: $\mathcal O(NM)$ $O (N M)$

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