Given a square matrix of size N*N, where each cell is associated with a specific cost. A path is defined as a specific sequence of cells which starts from top left cell move only right or down and ends on bottom right cell. We want to find a path with maximum average over all existing paths. Average is computed as total cost divided by number of cells visited in path.
Input : Matrix = [1, 2, 3 4, 5, 6 7, 8, 9] Output : 5.8 Path with maximum average is, 1 -> 4 -> 7 -> 8 -> 9 Sum of the path is 29 and average is 29/5 = 5.8
One interesting observation is, the only allowed moves are down and right, we need N-1 down moves and N-1 right moves to reach destination (bottom rightmost). So any path from from top left corner to bottom right corner requires 2N – 1 cells. In average value, denominator is fixed and we need to just maximize numerator. Therefore we basically need to to find maximum sum path. Calculating maximum sum of path is a classic dynamic programming problem, if dp[i][j] represents maximum sum till cell (i, j) from (0, 0) then at each cell (i, j), we update dp[i][j] as below,
for all i, 1 <= i <= N dp[i] = dp[i-1] + cost[i]; for all j, 1 <= j <= j dp[j] = dp[j-1] + cost[j]; otherwise dp[i][j] = max(dp[i-1][j], dp[i][j-1]) + cost[i][j];
Once we get maximum sum of all paths we will divide this sum by (2N – 1) and we will get our maximum average.
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Improved By : nitin mittal