Add Solution1.cpp for Range Queries Difference Array

algos/range_queries/DifferenceArray/soln/Chithra/Solution1.cpp

@@ -0,0 +1,68 @@
+/*
+Problem Statement:
+------------------
+Given an array of integers, perform multiple range update queries. 
+Each query is of the form [L, R, X] where you need to add X to all elements from index L to R (inclusive). 
+Return the final updated array.
+
+Approach:
+---------
+We use a Difference Array to perform range updates efficiently in O(1) time per query.
+
+Steps:
+1. Create a difference array `diff` of size n+1.
+2. For each query [L, R, X]:
+   - diff[L] += X
+   - diff[R+1] -= X
+3. Construct the final array by taking the prefix sum of `diff` and adding it to the original array.
+
+Time Complexity:
+----------------
+- O(N + Q), where N = size of the array, Q = number of queries
+
+Space Complexity:
+-----------------
+- O(N) for the difference array
+
+Example I/O:
+------------
+Input: arr = [1,2,3,4,5], queries = [[1,3,2],[0,2,1]]
+Output: [2,5,6,6,5]
+Explanation:
+- First query adds 2 to indices 1 to 3 => [1,4,5,6,5]
+- Second query adds 1 to indices 0 to 2 => [2,5,6,6,5]
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    int n = 5;
+    vector<int> arr = {1, 2, 3, 4, 5};
+    int queries[][3] = {{1, 3, 2}, {0, 2, 1}};
+    int q = 2;
+
+    vector<int> diff(n + 1, 0);
+
+    // Apply each query using difference array
+    for(int i = 0; i < q; i++) {
+        int l = queries[i][0];
+        int r = queries[i][1];
+        int val = queries[i][2];
+        diff[l] += val;
+        if(r + 1 < n) diff[r + 1] -= val;
+    }
+
+    // Build the final array using prefix sum
+    arr[0] += diff[0];
+    for(int i = 1; i < n; i++) {
+        diff[i] += diff[i-1];
+        arr[i] += diff[i];
+    }
+
+    // Print result
+    for(int x : arr) cout << x << " ";
+    cout << endl;
+
+    return 0;
+}