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Diffstat (limited to 'other/min_max_diff_pair.cpp')
-rw-r--r-- | other/min_max_diff_pair.cpp | 47 |
1 files changed, 47 insertions, 0 deletions
diff --git a/other/min_max_diff_pair.cpp b/other/min_max_diff_pair.cpp new file mode 100644 index 0000000..ac6dc3b --- /dev/null +++ b/other/min_max_diff_pair.cpp @@ -0,0 +1,47 @@ +// 2616. Minimize the Maximum Difference of Pairs +#include "leetcode.h" + +/* + * given a 0-indexed integer array 'nums' and an integer 'p'. find 'p' pairs of + * indices of 'nums' such that the maximum difference amongst all the pairs is + * minimised. also, ensure no index appears more than once amongst that 'p' + * pairs. note that for a pair of elements at the index 'i' and 'j'. the + * difference of this pair is '|nums[i] - nums[j]|', where '|x|' represents the + * absolute value of 'x'. return the minimum maximum difference among all 'p' + * pairs. we define the maximum of an empty set to be zero. + */ + +class Solution { + bool can_form_pair(vector<int> &nums, int p, int mid) { + int cnt = 0; + for (int i = 0; i < nums.size() - 1 && cnt < p;) { + if (nums[i + 1] - nums[i] <= mid) { + cnt++; + i += 2; + } else + i++; + } + return cnt >= p; + } + +public: + int minimizeMax(vector<int> &nums, int p) { + sort(nums.begin(), nums.end()); + int left = 0, right = nums.end() - nums.begin(); + while (left < right) { + int mid = (left + right) / 2; + if (can_form_pair(nums, p, mid)) + right = mid; + else + left = mid + 1; + } + return left; + } +}; + +int main() { + Solution obj; + vector<int> n1 = {10, 1, 2, 7, 1, 3}, n2 = {4, 2, 1, 2}; + printf("%d\n", obj.minimizeMax(n1, 2)); // expect: 1 + printf("%d\n", obj.minimizeMax(n2, 1)); // expect: 0 +} |