Given an array arr[] consisting of N integers, the task is to find the number of pairs (i, j) whose sum of indices is the same as the sum elements at the indices.
Examples:
Input: arr[] = {0, 1, 7, 4, 3, 2}
Output: 1
Explanation: There exists only pair that satisfies the condition is {(0, 1)}.Input: arr[] = {1, 6, 2, 4, 5, 6}
Output: 0
Naive Approach: The simple approach to solve the given problem is to generate all possible pairs of the given array and if the sum of any pairs is the same as the sum of its indices, then count this pair. After checking for all the pairs, print the total count obtained.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find all possible pairs// of the given array such that the sum// of arr[i] + arr[j] is i + jvoid countPairs(int arr[], int N){ // Stores the total count of pairs int answer = 0; // Iterate over the range for (int i = 0; i < N; i++) { // Iterate over the range for (int j = i + 1; j < N; j++) { if (arr[i] + arr[j] == i + j) { answer++; } } } // Print the total count cout << answer;}// Driver Codeint main(){ int arr[] = { 0, 1, 2, 3, 4, 5 }; int N = sizeof(arr) / sizeof(arr[0]); countPairs(arr, N); return 0;} |
Java
// Java program for the above approachimport java.io.*;public class GFG{// Function to find all possible pairs// of the given array such that the sum// of arr[i] + arr[j] is i + jpublic static void countPairs(int arr[], int N){ // Stores the total count of pairs int answer = 0; // Iterate over the range for(int i = 0; i < N; i++) { // Iterate over the range for(int j = i + 1; j < N; j++) { if (arr[i] + arr[j] == i + j) { answer++; } } } // Print the total count System.out.println(answer);}// Driver Codepublic static void main(String args[]){ int arr[] = { 0, 1, 2, 3, 4, 5 }; int N = arr.length; countPairs(arr, N);}}// This code is contributed by gfgking |
Python3
# Python3 program for the above approach # Function to find all possible pairs # of the given array such that the sum # of arr[i] + arr[j] is i + j def countPairs(arr, N): # Stores the total count of pairs answer = 0 # Iterate over the range for i in range(N): # Iterate over the range for j in range(i + 1, N): if arr[i] + arr[j] == i + j: answer += 1 # Print the total count print(answer)# Driver codearr = [ 0, 1, 2, 3, 4, 5 ]N = len(arr)countPairs(arr, N)# This code is contributed by Parth Manchanda |
C#
// C# program for the above approachusing System;class GFG{// Function to find all possible pairs// of the given array such that the sum// of arr[i] + arr[j] is i + jstatic void countPairs(int[] arr, int N){ // Stores the total count of pairs int answer = 0; // Iterate over the range for(int i = 0; i < N; i++) { // Iterate over the range for(int j = i + 1; j < N; j++) { if (arr[i] + arr[j] == i + j) { answer++; } } } // Print the total count Console.Write(answer);}// Driver Codepublic static void Main(string[] args){ int[] arr = { 0, 1, 2, 3, 4, 5 }; int N = arr.Length; countPairs(arr, N);}}// This code is contributed by target_2 |
Javascript
<script> // JavaScript program for the above approach // Function to find all possible pairs // of the given array such that the sum // of arr[i] + arr[j] is i + j function countPairs(arr, N) { // Stores the total count of pairs let answer = 0; // Iterate over the range for (let i = 0; i < N; i++) { // Iterate over the range for (let j = i + 1; j < N; j++) { if (arr[i] + arr[j] == i + j) { answer++; } } } // Print the total count document.write(answer); } // Driver Code let arr = [0, 1, 2, 3, 4, 5]; let N = arr.length; countPairs(arr, N); // This code is contributed by Potta Lokesh </script> |
Output:
15
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized by using an unordered map to store the count of elements having (arr[i] – i) value in the array arr[]. Follow the steps below to solve the problem:
- Initialize the variable, say answer as 0 to store the count of pairs in the array arr[].
- Initialize an unordered map mp[] to store the frequency of an element in the array arr[] having value (arr[i] – i).
- Iterate over the range [0, N] using the variable i and perform the following steps:
- Initialize the variable keyValue as the value of (arr[i] – i).
- Increase the value of keyValue in the unordered map mp[] by 1.
- Iterate over the unordered map mp[] using the variable i and perform the following steps:
- Initialize the variable size as i.second the value of the unordered map mp[].
- Add the value of size*(size – 1)/2 to the variable answer.
- After performing the above steps, print the value of the answer as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find all possible pairs// of the given array such that the sum// of arr[i] + arr[j] is i + jvoid countPairs(int arr[], int N){ // Stores the total count of pairs int answer = 0; unordered_map<int, int> mp; // Iterate over the range [0, N] for (int i = 0; i < N; i++) { int keyValue = arr[i] - i; mp[keyValue]++; } // Iterate over the range [0, N] for (auto i : mp) { int size = i.second; answer += (size * (size - 1)) / 2; } // Print the answer cout << answer;}// Driver Codeint main(){ int arr[] = { 0, 1, 2, 3, 4, 5 }; int N = sizeof(arr) / sizeof(arr[0]); countPairs(arr, N); return 0;} |
Java
/*package whatever //do not write package name here */// Java program for the above approachimport java.io.*;import java.util.*;class GFG { // Function to find all possible pairs// of the given array such that the sum// of arr[i] + arr[j] is i + j public static void countPairs(int[] arr, int n) { // Stores the total count of pairs int answer = 0; HashMap<Integer, Integer> mp = new HashMap<Integer, Integer>(); // Iterate over the range [0, N] for (int i = 0; i < n; i++) { int value = arr[i] - i; if (mp.containsKey(value)) { mp.put(value, mp.get(value) + 1); } else { mp.put(value, 1); } } // Iterate over the range [0, N] for (Map.Entry<Integer, Integer> map : mp.entrySet()) { int temp = map.getValue(); answer += temp * (temp - 1) / 2; } // Print the answer System.out.println(answer); } // Driver code public static void main(String[] args) { int[] arr = { 0, 1, 2, 3, 4, 5 }; int n = 6; countPairs(arr, n); }}// This code is contributed by maddler. |
Python3
# Python3 program for the above approach # Function to find all possible pairs # of the given array such that the sum # of arr[i] + arr[j] is i + j def countPairs(arr, N): # Stores the total count of pairs answer = 0 mp = {} # Iterate over the range [0, N] for i in range(N): keyValue = arr[i] - i if keyValue in mp.keys(): mp[keyValue] += 1 else: mp[keyValue] = 1 # Iterate over the range [0, N] for size in mp.values(): answer += (size * (size - 1)) // 2 print(answer)# Driver codearr = [ 0, 1, 2, 3, 4, 5 ]N = len(arr)countPairs(arr, N)# This code is contributed by Parth Manchanda |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function to find all possible pairs// of the given array such that the sum// of arr[i] + arr[j] is i + jstatic void countPairs(int []arr, int N){ // Stores the total count of pairs int answer = 0; Dictionary<int,int> mp = new Dictionary<int,int>(); // Iterate over the range [0, N] for (int i = 0; i < N; i++) { int keyValue = arr[i] - i; if(mp.ContainsKey(keyValue)) mp[keyValue]++; else mp.Add(keyValue,1); } // Iterate over the range [0, N] foreach(KeyValuePair<int,int> entry in mp) { int size = entry.Value; answer += (size * (size - 1)) / 2; } // Print the answer Console.Write(answer);}// Driver Codepublic static void Main(){ int []arr = {0, 1, 2, 3, 4, 5 }; int N = arr.Length; countPairs(arr, N);}}// This code is contributed by SURENDRA_GANGWAR. |
Javascript
<script>// Javascript program for the above approach// Function to find all possible pairs// of the given array such that the sum// of arr[i] + arr[j] is i + jfunction countPairs(arr, N) { // Stores the total count of pairs let answer = 0; let mp = new Map(); // Iterate over the range [0, N] for (let i = 0; i < N; i++) { let keyValue = arr[i] - i; if (mp.has(keyValue)) { mp.set(keyValue, mp.get(keyValue) + 1); } else { mp.set(keyValue, 1); } } // Iterate over the range [0, N] for (let i of mp) { let size = i[1]; answer += (size * (size - 1)) / 2; } // Print the answer document.write(answer);}// Driver Codelet arr = [0, 1, 2, 3, 4, 5];let N = arr.length;countPairs(arr, N);// This code is contributed by _saurabh_jaiswal.</script> |
Output:
15
Time Complexity: O(N)
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
