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// 1464. Maximum Product of Two Elements in an Array
#include "leetcode.h"
/*
* given the array of integers 'nums', you will choose two different indices 'i'
* and 'j' of thatarray. return the maximum value of '(nums[i] - 1) * (nums[j] -
* 1)'
*/
int maxProduct(int *nums, int nums_size) {
if (nums_size == 2)
return (nums[0] - 1) * (nums[1] - 1);
int max = nums[1] > nums[0] ? nums[1] : nums[0];
int min = nums[1] > nums[0] ? nums[0] : nums[1];
for (int i = 2; i < nums_size; i++) {
if (nums[i] > max) {
min = max;
max = nums[i];
} else if (nums[i] <= max && nums[i] > min)
min = nums[i];
}
return (max - 1) * (min - 1);
}
int main() {
int n1[] = {3, 4, 5, 2}, n2[] = {1, 5, 4, 5}, n3[] = {3, 7};
printf("%d\n", maxProduct(n1, ARRAY_SIZE(n1))); // expect: 12
printf("%d\n", maxProduct(n2, ARRAY_SIZE(n2))); // expect: 16
printf("%d\n", maxProduct(n3, ARRAY_SIZE(n3))); // expect: 12
}
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