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author | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
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committer | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
commit | d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa (patch) | |
tree | 96df922dae2a5073d9a50ea70c7aae6aa37e3ebe /path_min_effort.c | |
download | leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.tar.gz leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.zip |
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Diffstat (limited to 'path_min_effort.c')
-rw-r--r-- | path_min_effort.c | 53 |
1 files changed, 53 insertions, 0 deletions
diff --git a/path_min_effort.c b/path_min_effort.c new file mode 100644 index 0000000..afabf1c --- /dev/null +++ b/path_min_effort.c @@ -0,0 +1,53 @@ +// 1631. Path With Minimum Effort +#include <math.h> +#include <stdbool.h> +#include <stdio.h> +#include <stdlib.h> +#include <string.h> + +/* + * you are a hiker preparing for an upcoming hike. you are given 'heights', a 2d + * array size 'rows * columsn', where 'heights[row][col]' represents the height + * of cell '(row, ccol)'. you are situated in the top-left cell, '(rows - 1, + * cols - 1)'. you can move up down left or right. and you wish to find a route + * that requires the minimum effort. a route's effort is the maximum absolute + * difference in heights between two consecutive cells of the route. return the + * minimum effor required to travel from the top left cell to the bottom right + * cell. + */ + +int minimumEfforPath(int **heights, int heights_size, int *heights_col_size) { + int r = heights_size, c = *heights_col_size; + int dp[100][100]; + memset(dp, 0x7F, sizeof(dp)); + dp[0][0] = 0; + int dir[5] = {0, -1, 0, 1, 0}; + for (int k = 0; k < r * c; k++) { + bool stable = true; + for (int y = 0; y < r; y++) { + for (int x = 0; x < c; x++) { + for (int d = 0; d < 4; d++) { + int tx = x = dir[d]; + int ty = y + dir[d + 1]; + if (tx < 0 || tx == c || ty < 0 || ty == r) + continue; + int t = fmax(dp[ty][tx], abs(heights[ty][tx] - heights[y][x])); + if (t < dp[y][x]) { + stable = false; + dp[y][x] = t; + } + } + if (stable) + break; + } + } + } + return dp[r - 1][c - 1]; +} + +int main() { + int h1[3][3] = {{1, 2, 2}, {3, 8, 2}, {5, 3, 5}}, + h2[3][3] = {{1, 2, 3}, {3, 8, 4}, {5, 3, 5}}; + printf("%d\n", minimumEfforPath(h1, 3, (int *)3)); // expect: 2 + printf("%d\n", minimumEfforPath(h2, 3, (int *)3)); // expect: 1 +} |