diff options
Diffstat (limited to 'max_product_two_elem.c')
-rw-r--r-- | max_product_two_elem.c | 30 |
1 files changed, 30 insertions, 0 deletions
diff --git a/max_product_two_elem.c b/max_product_two_elem.c new file mode 100644 index 0000000..e52cf63 --- /dev/null +++ b/max_product_two_elem.c @@ -0,0 +1,30 @@ +// 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 +} |