Given an array **arr[]** of size **N**, the task is to count the number of pairs **(arr[i], arr[j])** such that **arr[j] – arr[i] = j – i**.

**Examples:**

Input:arr[] = {5, 2, 7}

Output:1

The only valid pair is (arr[0], arr[2]) as 7 – 5 = 2 – 0 = 2.

Input:arr[] = {1, 2, 3, 4}

Output:6

**Approach:** A pair **(arr[i], arr[j])** is said to be valid if **(arr[j] – arr[i]) = (j – i)**, it can also be written as **(arr[j] – j) = (arr[i] – i)** which is the difference of the element with its index. Now, the task is to divide the array into groups such that every group has equal difference of the element with its index then for every group if it has **N** elements, the count of possible pairs will be **(N * (N – 1)) / 2**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to return the count ` `// of all valid pairs ` `int` `countPairs(` `int` `arr[], ` `int` `n) ` `{ ` ` ` ` ` `// To store the frequencies ` ` ` `// of (arr[i] - i) ` ` ` `unordered_map<` `int` `, ` `int` `> map; ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `map[arr[i] - i]++; ` ` ` ` ` `// To store the required count ` ` ` `int` `res = 0; ` ` ` `for` `(` `auto` `x : map) { ` ` ` `int` `cnt = x.second; ` ` ` ` ` `// If cnt is the number of elements ` ` ` `// whose differecne with their index ` ` ` `// is same then ((cnt * (cnt - 1)) / 2) ` ` ` `// such pairs are possible ` ` ` `res += ((cnt * (cnt - 1)) / 2); ` ` ` `} ` ` ` ` ` `return` `res; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 1, 5, 6, 7, 9 }; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` ` ` `cout << countPairs(arr, n); ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the approach ` `import` `java.util.HashMap; ` ` ` `class` `GFG ` `{ ` ` ` ` ` `// Function to return the count ` ` ` `// of all valid pairs ` ` ` `static` `int` `countPairs(` `int` `arr[], ` `int` `n) ` ` ` `{ ` ` ` ` ` `// To store the frequencies ` ` ` `// of (arr[i] - i) ` ` ` `HashMap<Integer, ` ` ` `Integer> map = ` `new` `HashMap<Integer, ` ` ` `Integer>(); ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) ` ` ` `map.put(arr[i] - i, ` `0` `); ` ` ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) ` ` ` `map.put(arr[i] - i, map.get(arr[i] - i) + ` `1` `); ` ` ` ` ` `// To store the required count ` ` ` `int` `res = ` `0` `; ` ` ` `for` `(` `int` `x : map.values()) ` ` ` `{ ` ` ` `int` `cnt = x; ` ` ` ` ` `// If cnt is the number of elements ` ` ` `// whose differecne with their index ` ` ` `// is same then ((cnt * (cnt - 1)) / 2) ` ` ` `// such pairs are possible ` ` ` `res += ((cnt * (cnt - ` `1` `)) / ` `2` `); ` ` ` `} ` ` ` ` ` `return` `res; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main (String[] args) ` ` ` `{ ` ` ` `int` `arr[] = { ` `1` `, ` `5` `, ` `6` `, ` `7` `, ` `9` `}; ` ` ` `int` `n = arr.length; ` ` ` ` ` `System.out.println(countPairs(arr, n)); ` ` ` `} ` `} ` ` ` `// This code is contributed by AnkitRai01 ` |

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## Python3

`# Python3 implementation of the approach ` ` ` `# Function to return the count ` `# of all valid pairs ` `def` `countPairs(arr, n): ` ` ` ` ` `# To store the frequencies ` ` ` `# of (arr[i] - i) ` ` ` `map` `=` `dict` `() ` ` ` `for` `i ` `in` `range` `(n): ` ` ` `map` `[arr[i] ` `-` `i] ` `=` `map` `.get(arr[i] ` `-` `i, ` `0` `) ` `+` `1` ` ` ` ` `# To store the required count ` ` ` `res ` `=` `0` ` ` `for` `x ` `in` `map` `: ` ` ` `cnt ` `=` `map` `[x] ` ` ` ` ` `# If cnt is the number of elements ` ` ` `# whose differecne with their index ` ` ` `# is same then ((cnt * (cnt - 1)) / 2) ` ` ` `# such pairs are possible ` ` ` `res ` `+` `=` `((cnt ` `*` `(cnt ` `-` `1` `)) ` `/` `/` `2` `) ` ` ` ` ` `return` `res ` ` ` `# Driver code ` `arr ` `=` `[` `1` `, ` `5` `, ` `6` `, ` `7` `, ` `9` `] ` `n ` `=` `len` `(arr) ` ` ` `print` `(countPairs(arr, n)) ` ` ` `# This code is contributed by Mohit Kumar ` |

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## C#

`// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic; ` `class` `GFG ` `{ ` ` ` ` ` `// Function to return the count ` ` ` `// of all valid pairs ` ` ` `static` `int` `countPairs(` `int` `[]arr, ` `int` `n) ` ` ` `{ ` ` ` ` ` `// To store the frequencies ` ` ` `// of (arr[i] - i) ` ` ` `Dictionary<` `int` `, ` ` ` `int` `> map = ` `new` `Dictionary<` `int` `, ` ` ` `int` `>(); ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `map[arr[i] - i] = 0; ` ` ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `map[arr[i] - i]++; ` ` ` ` ` `// To store the required count ` ` ` `int` `res = 0; ` ` ` `foreach` `(KeyValuePair<` `int` `, ` `int` `> x ` `in` `map) ` ` ` `{ ` ` ` `int` `cnt = x.Value; ` ` ` ` ` `// If cnt is the number of elements ` ` ` `// whose differecne with their index ` ` ` `// is same then ((cnt * (cnt - 1)) / 2) ` ` ` `// such pairs are possible ` ` ` `res += ((cnt * (cnt - 1)) / 2); ` ` ` `} ` ` ` `return` `res; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `Main (String []args) ` ` ` `{ ` ` ` `int` `[]arr = { 1, 5, 6, 7, 9 }; ` ` ` `int` `n = arr.Length; ` ` ` ` ` `Console.WriteLine(countPairs(arr, n)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Arnab Kundu ` |

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**Output:**

3

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