Added the solution for prefix-sum

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

@@ -0,0 +1,47 @@
+/*
+Problem Statement:
+Given an array of integers, efficiently answer multiple range sum queries.
+Each query asks for the sum of elements between indices L and R (inclusive).
+Approach:
+We use a Prefix Sum array to answer range sum queries in O(1) time:
+1. Construct a prefix sum array `prefix` of size n:
+   - prefix[0] = arr[0]
+   - prefix[i] = prefix[i-1] + arr[i] for i = 1 to n-1
+2. For a query [L, R], the sum is:
+   - If L == 0: sum = prefix[R]
+   - Else: sum = prefix[R] - prefix[L-1]
+Time Complexity:
+- O(N) to build the prefix sum array
+- O(1) per query
+- Overall: O(N + Q), where Q = number of queries
+Space Complexity:
+- O(N) for the prefix sum array
+Example:
+Input: arr = [1,2,3,4,5], queries = [[1,3],[0,2]]
+Output: [9,6]
+Explanation:
+- Query [1,3] sum = 2+3+4 = 9
+- Query [0,2] sum = 1+2+3 = 6
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+int main() {
+    int n = 5;
+    vector<int> arr = {1, 2, 3, 4, 5};
+    int queries[][2] = {{1, 3}, {0, 2}};
+    int q = 2;
+    vector<int> prefix(n);
+    prefix[0] = arr[0];
+    for(int i = 1; i < n; i++) {
+        prefix[i] = prefix[i-1] + arr[i];
+    }
+    for(int i = 0; i < q; i++) {
+        int L = queries[i][0];
+        int R = queries[i][1];
+        int sum = (L == 0) ? prefix[R] : prefix[R] - prefix[L-1];
+        cout << sum << " ";
+    }
+    cout << endl;
+    return 0;
+}