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-rw-r--r--make_xor_segment_zero.c45
-rw-r--r--make_xor_segment_zero.py41
2 files changed, 86 insertions, 0 deletions
diff --git a/make_xor_segment_zero.c b/make_xor_segment_zero.c
new file mode 100644
index 0000000..9dbba8f
--- /dev/null
+++ b/make_xor_segment_zero.c
@@ -0,0 +1,45 @@
+// 1787. Make the XOR of All Segments Equal to Zero
+#include "leetcode.h"
+
+/*
+ * given an array 'nums' and an integer 'k'. the xor of a segment '[left,
+ * right]' where 'left <= right' is the xor of all the elements with indices
+ * between 'left' and 'right', inclusive 'nums[left] xor nums[left + 1] xor ...
+ * xor nums[right]'. return the minimum number of elements to change in the
+ * array such that the 'xor' of all segments of size 'k' is equal to zero
+ */
+
+int minChanges(int *nums, int numsSize, int k) {
+ int freq[k][1025], dp[k + 1][1025];
+ memset(freq, 0, sizeof(freq));
+ for (int i = 0; i <= k; i++)
+ for (int j = 0; j < 1025; j++)
+ dp[i][j] = 1e9 + 5;
+ dp[0][0] = 0;
+ for (int i = 0; i < numsSize; i++)
+ freq[i % k][nums[i]]++;
+ int prev_min = 0, prev_min2 = 1e9 + 5;
+ for (int i = 0; i < k; i++) {
+ for (int j = 0; j < 1025; j++)
+ for (int l = 0; l < numsSize; l += k)
+ if ((j ^ nums[l]) < 1025)
+ dp[i + 1][j ^ nums[l]] = fmin(
+ dp[i + 1][j ^ nums[l]],
+ dp[i][j] + ceil((numsSize - i) * 1.0 / k) - freq[i][nums[l]]);
+ for (int j = 0; j < 1025; j++) {
+ dp[i + 1][j] =
+ fmin(dp[i + 1][j], ceil((numsSize - i) * 1.0 / k) + prev_min);
+ prev_min2 = fmin(prev_min2, dp[i + 1][j]);
+ }
+ prev_min = prev_min2;
+ }
+ return dp[k][0];
+}
+
+int main() {
+ int n1[] = {1, 2, 0, 3, 0}, n2[] = {3, 4, 5, 2, 1, 7, 3, 4, 7},
+ n3[] = {1, 2, 4, 1, 2, 5, 1, 2, 6};
+ printf("%d\n", minChanges(n1, ARRAY_SIZE(n1), 1)); // expect: 3
+ printf("%d\n", minChanges(n2, ARRAY_SIZE(n2), 3)); // expect: 3
+ printf("%d\n", minChanges(n3, ARRAY_SIZE(n3), 3)); // expect: 3
+}
diff --git a/make_xor_segment_zero.py b/make_xor_segment_zero.py
new file mode 100644
index 0000000..2174968
--- /dev/null
+++ b/make_xor_segment_zero.py
@@ -0,0 +1,41 @@
+# 1787. Make the XOR of All Segments Equal to Zero
+import collections
+
+"""
+given an array 'nums' and an integer 'k'. the xor of a segment '[left,
+right]' where 'left <= right' is the xor of all the elements with indices
+between 'left' and 'right', inclusive 'nums[left] xor nums[left + 1] xor ...
+xor nums[right]'. return the minimum number of elements to change in the
+array such that the 'xor' of all segments of size 'k' is equal to zero
+"""
+
+
+class Solution(object):
+ def minChanges(self, nums, k):
+ """
+ :type nums: List[int]
+ :type k: int
+ :rtype: int
+ """
+ m = 1 << 10
+ freq = collections.defaultdict(collections.Counter)
+ for i, n in enumerate(nums):
+ freq[i % k][n] += 1
+ dp = [[0] * m for _ in range(k + 1)]
+ for i in range(1, m):
+ dp[0][i] = -float("inf")
+ for i in range(1, k + 1):
+ max_row = max(dp[i - 1])
+ for j in range(m):
+ for mask in freq[i - 1]:
+ dp[i][j] = max(
+ dp[i][j], max_row, dp[i - 1][j ^ mask] + freq[i - 1][mask]
+ )
+ return len(nums) - dp[-1][0]
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.minChanges(nums=[1, 2, 0, 3, 0], k=1))
+ print(obj.minChanges(nums=[3, 4, 5, 2, 1, 7, 3, 4, 7], k=3))
+ print(obj.minChanges(nums=[1, 2, 4, 1, 2, 5, 1, 2, 6], k=3))