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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
}
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