diff options
Diffstat (limited to 'max_sum_pair_equal_digits.c')
-rw-r--r-- | max_sum_pair_equal_digits.c | 32 |
1 files changed, 32 insertions, 0 deletions
diff --git a/max_sum_pair_equal_digits.c b/max_sum_pair_equal_digits.c new file mode 100644 index 0000000..ecf2bcf --- /dev/null +++ b/max_sum_pair_equal_digits.c @@ -0,0 +1,32 @@ +// 2342. Max Sum of a Pair With Equal Sum of Digits +#include "leetcode.h" + +/* + * you are given a 0-indexed array 'nums' consisting of positive integers. you + * can choose two indices 'i' and 'j', such that 'i != j', and the sum of the + * digits of the number 'nums[i]' is equal to that of 'nums[j]'. return the + * maximum value of 'nums[i] + nums[j]' that you can obtain over all + * possible indices 'i' and 'j' that satisfy the conditions. + */ + +int maximumSum(int *nums, int numsSize) { + int map[82], ans = -1; + memset(map, -1, sizeof(map)); + for (int i = 0; i < numsSize; i++) { + int sum = 0, tmap = nums[i]; + while (tmap) { + sum += tmap % 10; + tmap /= 10; + } + if (map[sum] != -1) + ans = fmax(ans, nums[i] + map[sum]); + map[sum] = fmax(map[sum], nums[i]); + } + return ans; +} + +int main() { + int n1[] = {18, 43, 36, 13, 7}, n2[] = {10, 12, 19, 14}; + printf("%d\n", maximumSum(n1, ARRAY_SIZE(n1))); // expect: 54 + printf("%d\n", maximumSum(n2, ARRAY_SIZE(n2))); // expect: -1 +} |