Create 1590. Make Sum Divisible by P 1 (#945)

Author: Chayandas07

1590. Make Sum Divisible by P 1

@@ -0,0 +1,37 @@
+class Solution {
+public:
+    int minSubarray(vector<int>& nums, int p) {
+        long long sufsum = 0;
+        for (auto x : nums) {
+            sufsum += x;                  
+        }
+
+        long long presum = 0;
+        int n = nums.size();
+        int ans = n;
+        unordered_map<int, int> dp;
+
+        // "Virtual" prefix: sum 0 at index -1 (empty prefix)
+        dp[0] = -1;
+
+        for (int i = 0; i < n; i++) {
+            presum += nums[i];            // extend prefix
+            sufsum -= nums[i];            // shrink suffix
+
+            // Store the latest index with this prefix remainder
+            dp[presum % p] = i;
+
+            // Right remainder and required left remainder
+            int rem = (p - sufsum % p) % p;
+
+            // If we have some prefix with this remainder,
+            // try removing between it and i
+            if (dp.find(rem) != dp.end()) {
+                ans = min(ans, i - dp[rem]);
+            }
+        }
+
+        if (ans == n) return -1;
+        return ans;
+    }
+};