82 lines
1.8 KiB
Java
82 lines
1.8 KiB
Java
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package DynamicProgramming;
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/*
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Given the following grid with length m and width n:
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\---\---\---\ (n)
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\ 1 \ 3 \ 1 \
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\---\---\---\
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\ 1 \ 5 \ 1 \
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\---\---\---\
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\ 4 \ 2 \ 1 \
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\---\---\---\
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(m)
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Find the path where its sum is the smallest.
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All numbers given are positive.
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The Time Complexity of your algorithm should be smaller than or equal to O(mn).
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The Space Complexity of your algorithm should be smaller than or equal to O(mn).
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You can only move from the top left corner to the down right corner.
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You can only move one step down or right.
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EXAMPLE:
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INPUT: grid = [[1,3,1],[1,5,1],[4,2,1]]
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OUTPUT: 7
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EXPLANATIONS: 1 + 3 + 1 + 1 + 1 = 7
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For more information see https://www.geeksforgeeks.org/maximum-path-sum-matrix/
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*/
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public class MinimumPathSum {
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public void testRegular() {
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int[][] grid = {
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{1, 3, 1},
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{1, 5, 1},
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{4, 2, 1}
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};
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System.out.println(minimumPathSum(grid));
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}
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public void testLessColumns() {
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int[][] grid = {
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{1, 2},
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{5, 6},
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{1, 1}
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};
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System.out.println(minimumPathSum(grid));
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}
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public void testLessRows() {
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int[][] grid = {
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{2, 3, 3},
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{7, 2, 1}
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};
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System.out.println(minimumPathSum(grid));
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}
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public void testOneRowOneColumn() {
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int[][] grid = {{2}};
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System.out.println(minimumPathSum(grid));
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}
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public static int minimumPathSum(int[][] grid) {
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int m = grid.length, n = grid[0].length;
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if (n == 0) {
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return 0;
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}
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int[][] dp = new int[m][n];
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dp[0][0] = grid[0][0];
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for (int i = 0; i < n - 1; i++) {
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dp[0][i + 1] = dp[0][i] + grid[0][i + 1];
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}
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for (int i = 0; i < m - 1; i++) {
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dp[i + 1][0] = dp[i][0] + grid[i + 1][0];
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}
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for (int i = 1; i < m; i++) {
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for (int j = 1; j < n; j++) {
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dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
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}
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}
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return dp[m - 1][n - 1];
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}
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}
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