Added Count Distinct Subarrays Divisible by K (LeetCode) solution with explanation and test cases

Author: Aditya12604

CPP/algorithms/arrays/Count_Distinct_Subarrays_Divisible_By_K.cpp

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+/*
+Problem Name: Count Distinct Subarrays Divisible by K in Sorted Array
+Problem Link: https://leetcode.com/problems/count-distinct-subarrays-divisible-by-k-in-sorted-array/
+Platform: LeetCode
+Language: C++
+Author: Aditya Shimoga
+Date: 30th October 2025
+
+Description:
+-----------------------------------
+Given a sorted array nums and an integer k, find the number of distinct subarrays
+whose sum is divisible by k.
+
+Key Insight:
+-----------------------------------
+- For any subarray [l, r], sum(l..r) is divisible by k ⇔ prefixSum[r] % k == prefixSum[l-1] % k.
+- To count valid subarrays efficiently, we maintain a frequency map of prefix sums modulo k.
+- Since the array is sorted and may contain repeated elements, we process contiguous "blocks"
+  of equal numbers separately. This ensures we don’t count identical subarrays multiple times.
+- For each block:
+  1. Calculate all valid subarrays ending in this block using prefix remainders from previous blocks.
+  2. Update the remainder frequency map after processing the block.
+
+Complexity:
+-----------------------------------
+Time Complexity: O(N)
+Space Complexity: O(K) , depending on distinct remainders
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+class Solution {
+public:
+    long long numGoodSubarrays(vector<int>& nums, int k) {
+        long long divisor = 1LL * k;
+        long long prefixSum = 0LL, resultCount = 0LL;
+        int n = nums.size();
+
+        map<int, int> remainderFreq;  // Tracks frequency of prefix sums modulo k
+
+        int i = 0;
+        while (i < n) {
+            int j = i;
+            int currentValue = nums[i];
+            long long tempPrefixSum = prefixSum;
+
+            // Step 1: Count valid subarrays ending within the current block
+            while (j < n && nums[j] == currentValue) {
+                prefixSum += nums[j];
+                int remainder = prefixSum % divisor;
+
+                // Subarray from start (remainder == 0)
+                if (remainder == 0) resultCount++;
+
+                // Subarray ending here matches an earlier prefix remainder
+                if (remainderFreq.count(remainder))
+                    resultCount += 1LL * remainderFreq[remainder];
+
+                j++;
+            }
+
+            // Step 2: Update frequency map for this block
+            j = i;
+            while (j < n && nums[j] == currentValue) {
+                tempPrefixSum += nums[j];
+                int remainder = tempPrefixSum % divisor;
+                remainderFreq[remainder]++;
+                j++;
+            }
+
+            // Move to next distinct value block
+            i = j;
+        }
+
+        return resultCount;
+    }
+};
+
+/*
+Example:
+Input:
+nums = [1,1,2,2,3], k = 3
+
+Output:
+Number of distinct subarrays divisible by 3 = 4
+
+Explanation:
+Subarrays divisible 3: [3], [1,2], [1,1,2,2],[1,1,2,2,3]
+*/
+
+int main() {
+    Solution solver;
+
+    // Test Case 1
+    vector<int> nums1 = {1, 1, 2, 2, 3};
+    int k1 = 3;
+    long long result1 = solver.numGoodSubarrays(nums1, k1);
+    cout << "Test Case 1:" << endl;
+    cout << "Input: nums = [1,1,2,2,3], k = 3" << endl;
+    cout << "Output: " << result1 << endl;
+    cout << "Expected: 4" << endl << endl;
+
+    // Test Case 2
+    vector<int> nums2 = {2, 2, 2, 4, 4};
+    int k2 = 4;
+    long long result2 = solver.numGoodSubarrays(nums2, k2);
+    cout << "Test Case 2:" << endl;
+    cout << "Input: nums = [2,2,2,4,4], k = 4" << endl;
+    cout << "Output: " << result2 << endl;
+    cout << "Expected: 6" << endl;
+
+    return 0;
+}