Added Difference Array solution for Issue #109

algos/rangeQueries/differenceArray/soln/Kiran95021/Solution1.cpp

@@ -0,0 +1,55 @@
+/*
+ * Problem: Range Addition (Standard Easy Problem)
+ *
+ * Problem Statement:
+ * Given an integer N (size of array) and Q (number of queries).
+ * Initially, the array has all zeros.
+ * Each query consists of three integers L, R, and Val.
+ * For each query, add 'Val' to all elements in the range [L, R] (1-based indexing).
+ * Finally, print the resulting array.
+ *
+ * Approach:
+ * We use the Difference Array technique to handle range updates in O(1).
+ * 1. Create a difference array 'diff' of size N+2 initialized to 0.
+ * 2. For each update (L, R, Val), perform:
+ * diff[L] += Val;
+ * diff[R+1] -= Val;
+ * 3. Compute the prefix sum of the 'diff' array to get the final values.
+ *
+ * Time Complexity: O(N + Q)
+ * Space Complexity: O(N)
+ */
+
+#include <iostream>
+#include <vector>
+
+using namespace std;
+
+void solve() {
+    int N, Q;
+    if (!(cin >> N >> Q)) return;
+
+    // 1-based indexing, so size is N+2
+    vector<long long> diff(N + 2, 0);
+
+    for (int i = 0; i < Q; i++) {
+        int L, R, Val;
+        cin >> L >> R >> Val;
+        diff[L] += Val;
+        diff[R + 1] -= Val;
+    }
+
+    long long currentVal = 0;
+    for (int i = 1; i <= N; i++) {
+        currentVal += diff[i];
+        cout << currentVal << " ";
+    }
+    cout << endl;
+}
+
+int main() {
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+    solve();
+    return 0;
+}