Solution added for 'Make Sum Divisible by P' Leetcode Problem 1590

Java/sum_divisible_by_p.java

@@ -0,0 +1,53 @@
+import java.util.HashMap;
+
+public class sum_divisible_by_p {
+    public static int minSubarray(int[] arr, int divisor) {
+        long totalSum = 0;
+        for (int num : arr) {
+            totalSum += num;
+        }
+        int remainder = (int)(totalSum % divisor);
+
+        // If the remainder is 0, the total sum is already divisible by divisor
+        if (remainder == 0) return 0;
+
+        // HashMap to store moduli of prefix sums and their indices
+        HashMap<Integer, Integer> prefixModMap = new HashMap<>();
+        prefixModMap.put(0, -1);
+
+        long cumulativeSum = 0;
+        int minLength = arr.length;
+
+        for (int i = 0; i < arr.length; ++i) {
+            cumulativeSum += arr[i];
+            int currentMod = (int)(cumulativeSum % divisor);
+
+            // Calculate the modulus we're looking for to find a subarray that, when removed, makes the sum divisible by divisor
+            int targetMod = (currentMod - remainder + divisor) % divisor;
+
+            // If the target modulus exists in the map, we found a subarray
+            if (prefixModMap.containsKey(targetMod)) {
+                minLength = Math.min(minLength, i - prefixModMap.get(targetMod));
+            }
+            prefixModMap.put(currentMod, i);
+        }
+        return minLength == arr.length ? -1 : minLength;
+    }
+
+    public static void main(String args[]){
+        int ans = minSubarray(new int[]{3, 1, 4, 2}, 6);
+        System.out.println(ans);
+    }
+}
+
+/*
+Demo Outputs:
+Input: nums = [3,1,4,2], p = 6
+Output: 1
+
+Input: nums = [6,3,5,2], p = 9
+Output: 2
+
+Input: nums = [1,2,3], p = 3
+Output: 0
+*/