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-rw-r--r-- | make_xor_segment_zero.c | 45 | ||||
-rw-r--r-- | make_xor_segment_zero.py | 41 |
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)) |