add 1658

Author: luliyucoordinate

README.md

@@ -765,6 +765,7 @@ LeetCode
 |1444|[Number of Ways of Cutting a Pizza](https://leetcode.com/problems/number-of-ways-of-cutting-a-pizza/)||[c++](./src/1444-Number-of-Ways-of-Cutting-a-Pizza/1444.cpp)|[python](./src/1444-Number-of-Ways-of-Cutting-a-Pizza/1444.py)|[go](./src/1444-Number-of-Ways-of-Cutting-a-Pizza/1444.go)|[js](./src/1444-Number-of-Ways-of-Cutting-a-Pizza/1444.js)|[java](./src/1444-Number-of-Ways-of-Cutting-a-Pizza/1444.java)||Hard|
 |1546|[Maximum Number of Non-Overlapping Subarrays With Sum Equals Target](https://leetcode.com/problems/maximum-number-of-non-overlapping-subarrays-with-sum-equals-target/)||[c++](./src/1546-Maximum-Number-of-Non-Overlapping-Subarrays-With-Sum-Equals-Target/maximum_number_of_nonoverlapping_with_sum_equals_target.cpp)||||||Medium|
 |1657|[Determine if Two Strings Are Close](https://leetcode.com/problems/determine-if-two-strings-are-close/)||[c++](./src/1657-Determine-if-Two-Strings-Are-Close/1657.cpp)||||||Medium|
+|1658|[Minimum Operations to Reduce X to Zero](https://leetcode.com/problems/minimum-operations-to-reduce-x-to-zero/)||[c++](./src/1658-Minimum-Operations-to-Reduce-X-to-Zero/1658.cpp)||||||Medium|
 |1673|[Find the Most Competitive Subsequence](https://leetcode.com/problems/find-the-most-competitive-subsequence/)||[c++](./src/1673-Find-the-Most-Competitive-Subsequence/1673.cpp)||||||Medium|
 |1691|[Maximum Height by Stacking Cuboids](https://leetcode.com/problems/maximum-height-by-stacking-cuboids/description/)||[c++](./src/1691-Maximum-Height-by-Stacking-Cuboids/1691.cpp)||||||Hard|
 |1704|[Determine if String Halves Are Alike](https://leetcode.com/problems/determine-if-string-halves-are-alike/)||[c++](./src/1704-Determine-if-String-Halves-Are-Alike/1704.cpp)||||||Easy|

src/1658-Minimum-Operations-to-Reduce-X-to-Zero/1658.cpp

@@ -0,0 +1,14 @@
+class Solution {
+ public:
+  int minOperations(vector<int>& nums, int t) {
+    int x = accumulate(nums.begin(), nums.end(), 0) - t;
+    if (x < 0) return -1;
+    int res = -1, n = nums.size();
+    for (int i = 0, j = 0, cur = 0; j < n;) {
+      cur += nums[j++];
+      while (cur > x) cur -= nums[i++];
+      if (cur == x) res = max(res, j - i);
+    }
+    return res < 0 ? -1 : n - res;
+  }
+};