leetcode

commit bae85fca0fc54c49a3dc6e0434567f3e7c057fad
parent 74d9c08327a7e36a7df63018e572ebea58fa0362
Author: Jack Sangdahl <jack@pngu.org>
Date:   Wed, 19 Nov 2025 22:19:04 -0700

832. flipping an image

Diffstat:
A
flipping_an_image.c
 | 
58
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
A
flipping_an_image.py
 | 
27
+++++++++++++++++++++++++++

2 files changed, 85 insertions(+), 0 deletions(-)
diff --git a/flipping_an_image.c b/flipping_an_image.c
@@ -0,0 +1,58 @@
+// 832. Flipping an Image
+#include "leetcode.h"
+
+/*
+ * given an 'n x n' binary matrix 'image', flip the image horizontally, then
+ * invert it and return the resulting image. to flip an image horizontally means
+ * that each row of the image is reversed. to invert an image means that each 0
+ * is replaced by 1, and each 1 is replaced by 0.
+ */
+
+int **flipAndInvertImage(int **image, int imageSize, int *imageColSize,
+                         int *returnSize, int **returnColumnSizes) {
+  int n = imageSize, m = imageColSize[0], tmp = 0;
+  *returnColumnSizes = (int *)malloc(n * sizeof(int));
+  *returnSize = n;
+  for (int i = 0; i < n; i++)
+    (*returnColumnSizes)[i] = m;
+  for (int i = 0; i < n; i++)
+    for (int j = 0; j < m / 2; j++) {
+      tmp = image[i][n - 1 - j];
+      image[i][n - 1 - j] = image[i][j];
+      image[i][j] = tmp;
+    }
+  for (int i = 0; i < n; i++)
+    for (int j = 0; j < m; j++)
+      image[i][j] = !image[i][j];
+  return image;
+}
+
+int main() {
+  int i1i[3][3] = {{1, 1, 0}, {1, 0, 1}, {0, 0, 0}}, rs1, *rcs1;
+  int i2i[4][4] = {{1, 1, 0, 0}, {1, 0, 0, 1}, {0, 1, 1, 1}, {1, 0, 1, 0}}, rs2,
+      *rcs2;
+  struct two_d_arr *i1 =
+      two_d_arr_init(ARRAY_SIZE(i1i), ARRAY_SIZE(i1i[0]), i1i);
+  struct two_d_arr *i2 =
+      two_d_arr_init(ARRAY_SIZE(i2i), ARRAY_SIZE(i2i[0]), i2i);
+  int **faii1 =
+      flipAndInvertImage(i1->arr, i1->row_size, i1->col_size, &rs1, &rcs1);
+  int **faii2 =
+      flipAndInvertImage(i2->arr, i2->row_size, i2->col_size, &rs2, &rcs2);
+  for (int i = 0; i < rs1; i++) {
+    for (int j = 0; j < rcs1[i]; j++)
+      printf("%d ", faii1[i][j]);
+    printf("\n");
+  }
+  printf("\n");
+  for (int i = 0; i < rs2; i++) {
+    for (int j = 0; j < rcs2[i]; j++)
+      printf("%d ", faii2[i][j]);
+    printf("\n");
+  }
+  printf("\n");
+  free(rcs1);
+  two_d_arr_free(i1);
+  free(rcs2);
+  two_d_arr_free(i2);
+}
diff --git a/flipping_an_image.py b/flipping_an_image.py
@@ -0,0 +1,27 @@
+# 832. Flipping an Image
+
+"""
+given an 'n x n' binary matrix 'image', flip the image horizontally, then
+invert it and return the resulting image. to flip an image horizontally means
+that each row of the image is reversed. to invert an image means that each 0
+is replaced by 1, and each 1 is replaced by 0.
+"""
+
+
+class Solution(object):
+    def flipAndInvertImage(self, image):
+        """
+        :type image: List[List[int]]
+        :rtype: List[List[int]]
+        """
+        return [[1 ^ i for i in j[::-1]] for j in image]
+
+
+if __name__ == "__main__":
+    obj = Solution()
+    print(obj.flipAndInvertImage(image=[[1, 1, 0], [1, 0, 1], [0, 0, 0]]))
+    print(
+        obj.flipAndInvertImage(
+            image=[[1, 1, 0, 0], [1, 0, 0, 1], [0, 1, 1, 1], [1, 0, 1, 0]]
+        )
+    )