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Diffstat (limited to 'min_max_array.cpp')
-rw-r--r-- | min_max_array.cpp | 43 |
1 files changed, 0 insertions, 43 deletions
diff --git a/min_max_array.cpp b/min_max_array.cpp deleted file mode 100644 index 6da3f2b..0000000 --- a/min_max_array.cpp +++ /dev/null @@ -1,43 +0,0 @@ -// 2439. Minimize Maximum of Array -#include "leetcode.h" - -/* - * given a 0-indexed array 'nums' comprising of n unsigned integers. ine one - * operation you must choose an integer 'i' such that '1 <= i < n' and 'nums[i] - * > 0'. decrease 'nums[i]' by 1 or increase 'nums[i]' by 1. return minimum - * possible value of the max integer 'nums' after performing any number of - * operations. - */ - -class Solution { - bool valid(vector<int> &nums, int mid) { - long sum = 0; - for (int i = 0; i < nums.size(); i++) { - sum += nums[i]; - if (sum > (long)mid * (i + 1)) - return false; - } - return true; - } - -public: - int minimizeArrayValue(vector<int> &nums) { - int left = 0, right = *max_element(nums.begin(), nums.end()), ans = 0; - while (left <= right) { - int mid = (left + right) >> 1; - if (valid(nums, mid)) { - ans = mid; - right = mid - 1; - } else - left = mid + 1; - } - return ans; - } -}; - -int main() { - Solution obj; - vector<int> n1 = {3, 7, 1, 6}, n2 = {10, 1}; - printf("%d\n", obj.minimizeArrayValue(n1)); // expect: 5 - printf("%d\n", obj.minimizeArrayValue(n2)); // expect: 10 -} |