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+// 3022. Minimize OR of Remaining Elements Using Operations
+#include "leetcode.h"
+
+/*
+ * given a 0 indexed integer array 'nums' and an integer 'k'. in one operation,
+ * you can pick any index 'i' of 'nums' such that '0 <= i < nums.len - 1' and
+ * replace 'nums[i]' and 'nums[i + 1]' with a single occurence of 'nums[i] &
+ * nums[i + 1]' where & represents the bitwise and operation. return the minimum
+ * possible value of the bitwise or of the remaining elements of 'nums' after
+ * applying at most 'k' operations.
+ */
+
+int minOrAfterOperations(int *nums, int numsSize, int k) {
+ int max_elem = nums[0], v = (1 << 30) - 1, mask = 0;
+ for (int i = 1; i < numsSize; i++)
+ if (nums[i] > max_elem)
+ max_elem = nums[i];
+ for (int i = 29; i >= 0; --i) {
+ int curr = 1 << i;
+ mask |= curr;
+ if (curr > max_elem)
+ continue;
+ int tmp = numsSize;
+ bool vis = false;
+ for (int j = 0; j < numsSize && tmp > k; --tmp) {
+ int c = nums[i++];
+ for (; i < tmp && c & mask; c &= nums[i++])
+ ;
+ if (c & mask)
+ break;
+ vis = true;
+ }
+ if (!vis || tmp > k)
+ mask ^= curr;
+ }
+ return v ^ mask;
+}
+
+int main() {
+ int n1[] = {3, 5, 3, 2, 7}, n2[] = {7, 3, 15, 14, 2, 8},
+ n3[] = {10, 7, 10, 3, 9, 14, 9, 4};
+ printf("%d\n", minOrAfterOperations(n1, ARRAY_SIZE(n1), 2)); // expect: 3
+ printf("%d\n", minOrAfterOperations(n2, ARRAY_SIZE(n2), 4)); // expect: 2
+ printf("%d\n", minOrAfterOperations(n3, ARRAY_SIZE(n3), 15)); // expect: 15
+}