Given an array of integers, **arr[]** of size **N**, the task is to print all possible sum at each valid index, that can be obtained by adding **i + a[i]th** (*1-based indexing*) subsequent elements till **i ≤ N**.

**Examples:**

= {4, 1, 4}Input:arr[]Output:4 5 4Explanation:

For i = 1, arr[1] = 4.

For i = 2, arr[2] = arr[2] + arr[2 + arr[2]] = arr[2] + arr[2 + 1] = arr[2] + arr[3] = 1 + 4 = 5.

For i = 3, arr[3] = 4.

= {1, 2, 7, 1, 8}Input:arr[]12 11 7 9 8Output:Explanation:

For i = 1, arr[1] = arr[1] + arr[1 + 1] + arr[1 + 1 + 2] + arr[1 + 1 + 2 + 1] = arr[1] + arr[2] + arr[4] + arr[5] = 1 + 2 + 1 + 8 = 12.

For i = 2, arr[2] = arr[2] + arr[2 + 2] + arr[2 + 2 + 1] = 2 + 1 + 8 = 11.

For i = 3, arr[3] = 7.

For i = 4,arr[4] = arr[4] + arr[4 + 1] = 1 + 8 = 9.

For i = 5, the sum will be arr[5] = 8.

**Naive Approach:*** *The simplest approach is to traverse the array and for every i

^{th}index, keep updating

**arr[i]**to

**arr[i + arr[i]]**while

**i ≤ N**and print the sum at that index.

**Time Complexity: **O(N^{2})**Auxiliary Space: **O(N)

**Efficient Approach**: The above approach can be optimized by traversing the array in reverse and store the sum for every visited index to the current index. Finally, print the array.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to maximize value` `// at every array index by` `// performing given operations` `int` `maxSum(` `int` `arr[], ` `int` `N)` `{` ` ` `int` `ans = 0;` ` ` `// Traverse the array in reverse` ` ` `for` `(` `int` `i = N - 1; i >= 0; i--) {` ` ` `int` `t = i;` ` ` `// If the current index` ` ` `// is a valid index` ` ` `if` `(t + arr[i] < N) {` ` ` `arr[i] += arr[t + arr[i]];` ` ` `}` ` ` `}` ` ` `// Print the array` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `cout << arr[i] << ` `' '` `;` ` ` `}` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given array` ` ` `int` `arr[] = { 1, 2, 7, 1, 8 };` ` ` `// Size of the array` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `maxSum(arr, N);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.io.*;` `class` `GFG{` `// Function to maximize value` `// at every array index by` `// performing given operations` `static` `void` `maxSum(` `int` `[] arr, ` `int` `N)` `{` ` ` `int` `ans = ` `0` `;` ` ` `// Traverse the array in reverse` ` ` `for` `(` `int` `i = N - ` `1` `; i >= ` `0` `; i--)` ` ` `{` ` ` `int` `t = i;` ` ` `// If the current index` ` ` `// is a valid index` ` ` `if` `(t + arr[i] < N)` ` ` `{` ` ` `arr[i] += arr[t + arr[i]];` ` ` `}` ` ` `}` ` ` `// Print the array` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++)` ` ` `{` ` ` `System.out.print(arr[i] + ` `" "` `);` ` ` `}` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` ` ` `// Given array` ` ` `int` `[] arr = { ` `1` `, ` `2` `, ` `7` `, ` `1` `, ` `8` `};` ` ` `// Size of the array` ` ` `int` `N = arr.length;` ` ` `maxSum(arr, N);` `}` `}` `// This code is contributed by Dharanendra L V` |

## Python3

`# Python program for the above approach` `# Function to maximize value` `# at every array index by` `# performing given operations` `def` `maxSum(arr, N):` ` ` `ans ` `=` `0` `;` ` ` `# Traverse the array in reverse` ` ` `for` `i ` `in` `range` `(N ` `-` `1` `, ` `-` `1` `, ` `-` `1` `):` ` ` `t ` `=` `i;` ` ` `# If the current index` ` ` `# is a valid index` ` ` `if` `(t ` `+` `arr[i] < N):` ` ` `arr[i] ` `+` `=` `arr[t ` `+` `arr[i]];` ` ` `# Prthe array` ` ` `for` `i ` `in` `range` `(N):` ` ` `print` `(arr[i], end ` `=` `" "` `);` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `# Given array` ` ` `arr ` `=` `[` `1` `, ` `2` `, ` `7` `, ` `1` `, ` `8` `];` ` ` `# Size of the array` ` ` `N ` `=` `len` `(arr);` ` ` `maxSum(arr, N);` `# This code is contributed by 29AjayKumar` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG {` `// Function to maximize value` `// at every array index by` `// performing given operations` `static` `void` `maxSum(` `int` `[] arr, ` `int` `N)` `{` ` ` ` ` `// Traverse the array in reverse` ` ` `for` `(` `int` `i = N - 1; i >= 0; i--)` ` ` `{` ` ` `int` `t = i;` ` ` ` ` `// If the current index` ` ` `// is a valid index` ` ` `if` `(t + arr[i] < N)` ` ` `{` ` ` `arr[i] += arr[t + arr[i]];` ` ` `}` ` ` `}` ` ` `// Print the array` ` ` `for` `(` `int` `i = 0; i < N; i++)` ` ` `{` ` ` `Console.Write(arr[i] + ` `" "` `);` ` ` `}` `}` `// Driver Code` `static` `public` `void` `Main()` `{` ` ` ` ` `// Given array` ` ` `int` `[] arr = { 1, 2, 7, 1, 8 };` ` ` `// Size of the array` ` ` `int` `N = arr.Length;` ` ` `maxSum(arr, N);` `}` `}` `// This code is contributed by Dharanendra L V` |

## Javascript

`<script>` `// JavaScript program for` `// the above approach` ` ` `// Function to maximize value` `// at every array index by` `// performing given operations` `function` `maxSum(arr, N)` `{` ` ` `let ans = 0;` ` ` `// Traverse the array in reverse` ` ` `for` `(let i = N - 1; i >= 0; i--) {` ` ` `let t = i;` ` ` `// If the current index` ` ` `// is a valid index` ` ` `if` `(t + arr[i] < N) {` ` ` `arr[i] += arr[t + arr[i]];` ` ` `}` ` ` `}` ` ` `// Print the array` ` ` `for` `(let i = 0; i < N; i++) {` ` ` `document.write(arr[i] + ` `' '` `);` ` ` `}` `}` ` ` `// Driver Code` ` ` ` ` `// Given array` ` ` `let arr = [ 1, 2, 7, 1, 8 ];` ` ` `// Size of the array` ` ` `let N = arr.length;` ` ` `maxSum(arr, N);` ` ` `</script>` |

**Output:**

12 11 7 9 8

**Time Complexity: **O(N)**Auxiliary Space: **O(1)

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