From d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa Mon Sep 17 00:00:00 2001
From: jack <0x6A73@pm.me>
Date: Fri, 14 Jun 2024 13:13:15 -0600
Subject: initial

---
 path_min_effort.c | 53 +++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 53 insertions(+)
 create mode 100644 path_min_effort.c

(limited to 'path_min_effort.c')

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
+}
-- 
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