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Diffstat (limited to 'maximise_happiness_children.c')
-rw-r--r-- | maximise_happiness_children.c | 32 |
1 files changed, 32 insertions, 0 deletions
diff --git a/maximise_happiness_children.c b/maximise_happiness_children.c new file mode 100644 index 0000000..4199189 --- /dev/null +++ b/maximise_happiness_children.c @@ -0,0 +1,32 @@ +// 3075. Maximize Happiness of Selected Children +#include "leetcode.h" + +/* + * given an array 'happiness' of length 'n' and a positive integer 'k'. there + * are 'n' children standing in a queue where the i'th child has happiness value + * 'happiness[i]'. you want to select 'k' children from these 'n' children in + * 'k' turns. in each turn you select a child the happiness value of all the + * children that have not been selected till now decreases by one. note that the + * happiness value cannot become negative and decremented only if it is + * positive. return the maximum sum of the happiness values of the selected + * children you can achieve by selecting 'k' children. + */ + +int cmp(const void *a, const void *b) { return *(int *)b - *(int *)a; } + +long long maximumHappinessSum(int *happiness, int happinessSize, int k) { + long long ans = 0, sub = 0; + qsort(happiness, happinessSize, sizeof(int), cmp); + for (int i = 0; i < happinessSize && k; i++, k--) { + ans += happiness[i] - sub < 0 ? 0 : happiness[i] - sub; + sub++; + } + return ans; +} + +int main() { + int h1[] = {1, 2, 3}, h2[] = {1, 1, 1, 1}, h3[] = {2, 3, 4, 5}; + printf("%lld\n", maximumHappinessSum(h1, ARRAY_SIZE(h1), 2)); // expect: 4 + printf("%lld\n", maximumHappinessSum(h2, ARRAY_SIZE(h2), 2)); // expect: 1 + printf("%lld\n", maximumHappinessSum(h3, ARRAY_SIZE(h3), 1)); // expect: 5 +} |