# Recursive program to find all Indices of a Number

Given an array arr of size N and an integer X. The task is to find all the indices of the integer X in the array

Examples:

Input: arr = {1, 2, 3, 2, 2, 5}, X = 2
Output: 1 3 4
Element 2 is present at indices 1, 3, 4 (0 based indexing)

Input: arr[] = {7, 7, 7}, X = 7
Output: 0 1 2

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The iterative approach is simple, just traverse the given array and keep on storing the indices of the element in the another array.

Recursive approach:

• If the start index reaches the length of the array, then return empty array
• Else keep the first element of the array with yourself and pass the rest of the array to recursion.
• If the element at start index is not equal to x then just simply return the answer which came from recursion.
• Else if the element at start index is equal to x then shift the elements of the array (which is the answer of recursion) one step to the right and then put the start index in the front of the array (which came through recursion)

Below is the implementation of the above approach:

## C++

 `// CPP program to find all indices of a number ` `#include ` `using` `namespace` `std; ` ` `  `// A recursive function to find all ` `// indices of a number ` `int` `AllIndexesRecursive(``int` `input[], ``int` `size, ` `                    ``int` `x, ``int` `output[]) ` `{ ` `     `  `    ``// If an empty array comes ` `    ``// to the function, then ` `    ``// return zero ` `    ``if` `(size == 0) { ` `        ``return` `0; ` `    ``} ` ` `  `    ``// Getting the recursive answer ` `    ``int` `smallAns = AllIndexesRecursive(input + 1, ` `                                    ``size - 1, x, output); ` ` `  `    ``// If the element at index 0 is equal ` `    ``// to x then add 1 to the array values ` `    ``// and shift them right by 1 step ` `    ``if` `(input == x) { ` `        ``for` `(``int` `i = smallAns - 1; i >= 0; i--) { ` `            ``output[i + 1] = output[i] + 1; ` `        ``} ` ` `  `        ``// Put the start index in front ` `        ``// of the array ` `        ``output = 0; ` `        ``smallAns++; ` `    ``} ` `    ``else` `{ ` `         `  `        ``// If the element at index 0 is not equal ` `        ``// to x then add 1 to the array values ` `        ``for` `(``int` `i = smallAns - 1; i >= 0; i--) { ` `            ``output[i] = output[i] + 1; ` `        ``} ` `    ``} ` `    ``return` `smallAns; ` `} ` ` `  `// Function to find all indices of a number ` `void` `AllIndexes(``int` `input[], ``int` `n, ``int` `x) ` `{ ` `    ``int` `output[n]; ` `    ``int` `size = AllIndexesRecursive(input, n, ` `                                ``x, output); ` `    ``for` `(``int` `i = 0; i < size; i++) { ` `        ``cout << output[i] << ``" "``; ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3, 2, 2, 5 }, x = 2; ` `     `  `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `     `  `    ``// Function call ` `    ``AllIndexes(arr, n, x); ` `     `  `    ``return` `0; ` `} `

## Java

 `// Java program to find all ` `// indices of a number ` `public` `class` `GFG { ` ` `  `    ``public` `int``[] AllIndexesRecursive(``int` `input[], ` `                                ``int` `x, ``int` `start) ` `    ``{ ` `        ``// If the start index reaches the ` `        ``// length of the array, then ` `        ``// return empty array ` `        ``if` `(start == input.length) { ` `            ``int``[] ans = ``new` `int``[``0``]; ``// empty array ` `            ``return` `ans; ` `        ``} ` ` `  `        ``// Getting the recursive answer in ` `        ``// smallIndex array ` `        ``int``[] smallIndex = AllIndexesRecursive(input, x, ` `                                              ``start + ``1``); ` ` `  `        ``// If the element at start index is equal ` `        ``// to x then ` `        ``// (which is the answer of recursion) and then ` `        ``// (which came through recursion) ` `        ``if` `(input[start] == x) { ` `            ``int``[] myAns = ``new` `int``[smallIndex.length + ``1``]; ` ` `  `            ``// Put the start index in front ` `            ``// of the array ` `            ``myAns[``0``] = start; ` `            ``for` `(``int` `i = ``0``; i < smallIndex.length; i++) { ` `                 `  `                ``// Shift the elements of the array ` `                ``// one step to the right ` `                ``// and putting them in ` `                ``// myAns array ` `                ``myAns[i + ``1``] = smallIndex[i]; ` `            ``} ` `            ``return` `myAns; ` `        ``} ` `        ``else` `{ ` `             `  `            ``// If the element at start index is not ` `            ``// equal to x then just simply return the ` `            ``// answer which came from recursion. ` `            ``return` `smallIndex; ` `        ``} ` `    ``} ` ` `  `    ``public` `int``[] AllIndexes(``int` `input[], ``int` `x) ` `    ``{ ` ` `  `        ``return` `AllIndexesRecursive(input, x, ``0``); ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``GFG g = ``new` `GFG(); ` `        ``int` `arr[] = { ``1``, ``2``, ``3``, ``2``, ``2``, ``5` `}, x = ``2``; ` `         `  `        ``int` `output[] = g.AllIndexes(arr, x); ` `         `  `        ``// Printing the output array ` `        ``for` `(``int` `i = ``0``; i < output.length; i++) { ` `            ``System.out.print(output[i] + ``" "``); ` `        ``} ` `    ``} ` `} `

## Python3

 `# Python3 program to find all ` `# indices of a number ` `def` `AllIndexesRecursive(``input``, x, start): ` `     `  `    ``# If the start index reaches the ` `    ``# length of the array, then ` `    ``# return empty array ` `    ``if` `(start ``=``=` `len``(``input``)): ` `        ``ans ``=` `[] ``# empty array ` `        ``return` `ans ` ` `  `    ``# Getting the recursive answer in ` `    ``# smallIndex array ` `    ``smallIndex ``=` `AllIndexesRecursive(``input``, x,  ` `                                     ``start ``+` `1``) ` ` `  `    ``# If the element at start index is equal ` `    ``# to x then ` `    ``# (which is the answer of recursion) and then ` `    ``# (which came through recursion) ` `    ``if` `(``input``[start] ``=``=` `x): ` `        ``myAns ``=` `[``0` `for` `i ``in` `range``(``len``(smallIndex) ``+` `1``)] ` ` `  `        ``# Put the start index in front ` `        ``# of the array ` `        ``myAns[``0``] ``=` `start ` `        ``for` `i ``in` `range``(``len``(smallIndex)): ` ` `  `            ``# Shift the elements of the array ` `            ``# one step to the right ` `            ``# and putting them in ` `            ``# myAns array ` `            ``myAns[i ``+` `1``] ``=` `smallIndex[i] ` ` `  `        ``return` `myAns ` `    ``else``: ` ` `  `        ``# If the element at start index is not ` `        ``# equal to x then just simply return the ` `        ``# answer which came from recursion. ` `        ``return` `smallIndex ` ` `  `# Function to find all indices of a number ` `def` `AllIndexes(``input``, x): ` ` `  `    ``return` `AllIndexesRecursive(``input``, x, ``0``) ` ` `  `# Driver Code ` `arr ``=` `[ ``1``, ``2``, ``3``, ``2``, ``2``, ``5` `] ` `x ``=` `2` ` `  `output``=``AllIndexes(arr, x) ` ` `  `# Printing the output array ` `for` `i ``in` `output: ` `    ``print``(i, end ``=` `" "``) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# program to find all ` `// indices of a number ` `using` `System; ` `class` `GFG  ` `{ ` `    ``public` `int``[] AllIndexesRecursive(``int` `[]input, ` `                                        ``int` `x, ``int` `start) ` `    ``{ ` `        ``// If the start index reaches the ` `        ``// length of the array, then ` `        ``// return empty array ` `        ``if` `(start == input.Length)  ` `        ``{ ` `            ``int``[] ans = ``new` `int``; ``// empty array ` `            ``return` `ans; ` `        ``} ` ` `  `        ``// Getting the recursive answer in ` `        ``// smallIndex array ` `        ``int``[] smallIndex = AllIndexesRecursive(input, x, ` `                                               ``start + 1); ` ` `  `        ``// If the element at start index is equal ` `        ``// to x then ` `        ``// (which is the answer of recursion) and  ` `        ``// then (which came through recursion) ` `        ``if` `(input[start] == x) ` `        ``{ ` `            ``int``[] myAns = ``new` `int``[smallIndex.Length + 1]; ` ` `  `            ``// Put the start index in front ` `            ``// of the array ` `            ``myAns = start; ` `            ``for` `(``int` `i = 0; i < smallIndex.Length; i++) ` `            ``{ ` `                 `  `                ``// Shift the elements of the array ` `                ``// one step to the right ` `                ``// and putting them in ` `                ``// myAns array ` `                ``myAns[i + 1] = smallIndex[i]; ` `            ``} ` `            ``return` `myAns; ` `        ``} ` `        ``else`  `        ``{ ` `             `  `            ``// If the element at start index is not ` `            ``// equal to x then just simply return the ` `            ``// answer which came from recursion. ` `            ``return` `smallIndex; ` `        ``} ` `    ``} ` ` `  `    ``public` `int``[] AllIndexes(``int` `[]input, ``int` `x) ` `    ``{ ` `        ``return` `AllIndexesRecursive(input, x, 0); ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``GFG g = ``new` `GFG(); ` `        ``int` `[]arr = { 1, 2, 3, 2, 2, 5 }; ` `        ``int` `x = 2; ` `         `  `        ``int` `[]output = g.AllIndexes(arr, x); ` `         `  `        ``// Printing the output array ` `        ``for` `(``int` `i = 0; i < output.Length; i++)  ` `        ``{ ` `            ``Console.Write(output[i] + ``" "``); ` `        ``} ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67.. `

Output:

```1 3 4
```

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