From f2fdf2af71892342d1f8fc418d3d504dbb31ecf8 Mon Sep 17 00:00:00 2001
From: Kdev <120096635+KushagraSahu-01@users.noreply.github.com>
Date: Sun, 28 Dec 2025 01:17:53 +0530
Subject: [PATCH] Add solution for range queries using difference array

Implemented a solution for processing range update and point query using a difference array.
---
 .../soln/KushagraSahu/Soution1.cpp            | 56 +++++++++++++++++++
 1 file changed, 56 insertions(+)
 create mode 100644 algos/range_queries/DifferenceArray/soln/KushagraSahu/Soution1.cpp

diff --git a/algos/range_queries/DifferenceArray/soln/KushagraSahu/Soution1.cpp b/algos/range_queries/DifferenceArray/soln/KushagraSahu/Soution1.cpp
new file mode 100644
index 0000000..c756205
--- /dev/null
+++ b/algos/range_queries/DifferenceArray/soln/KushagraSahu/Soution1.cpp
@@ -0,0 +1,56 @@
+/*
+Given an array of n integers, your task is to process q queries of the following types:
+
+increase each value in range [a,b] by u
+what is the value at position k?
+
+Input
+The first input line has two integers n and q: the number of values and queries.
+The second line has n integers x_1,x_2,\dots,x_n: the array values.
+Finally, there are q lines describing the queries. Each line has three integers: either "1 a b u" or "2 k".
+Output
+Print the result of each query of type 2.
+
+Approach : we have to update the array multiple times so instead of running
+a loop everytime we create a difference array and then take the prefixSum of it when we require the final array
+Time Complexity : O(n) for calculating the prefix sum in the type 2 query
+Space Complexity : O(n) because of arr and diff arr of size n
+*/
+#include <iostream>
+using namespace std;
+int main()
+{
+    int n,q;
+    cin >>n>>q;
+    long long int arr[n],diff[n];
+    for (int i = 0; i < n; i++)
+    diff[i] = 0;    
+    for(int i=0;i<n;i++)
+    cin >>arr[i];
+    while(q--)
+    {
+        int type;
+        cin >>type;
+        if(type==1)
+        {
+            int a,b;
+            long long int u;
+            cin >>a>>b>>u;
+            a--;
+            b--;
+            diff[a]+=u;
+            if(b+1<n)
+            diff[b+1]-=u;
+        }
+        else
+        {
+            int k;
+            cin >>k;
+            k--;
+            long long sum=0;
+            for(int i=0;i<=k;i++)
+            sum+=diff[i];
+            cout <<arr[k]+sum<<"\n";
+        }
+    }
+}