Maximize every array element by repeatedly adding all valid i+a[i]th array element
Last Updated :
19 Jan, 2022
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:
Input: arr[] = {4, 1, 4}
Output: 4 5 4
Explanation:
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.
Input: arr[] = {1, 2, 7, 1, 8}
Output: 12 11 7 9 8
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 ith index, keep updating arr[i] to arr[i + arr[i]] while i ? N and print the sum at that index.
Time Complexity: O(N2)
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++
#include <bits/stdc++.h>
using namespace std;
int maxSum( int arr[], int N)
{
int ans = 0;
for ( int i = N - 1; i >= 0; i--) {
int t = i;
if (t + arr[i] < N) {
arr[i] += arr[t + arr[i]];
}
}
for ( int i = 0; i < N; i++) {
cout << arr[i] << ' ' ;
}
}
int main()
{
int arr[] = { 1, 2, 7, 1, 8 };
int N = sizeof (arr) / sizeof (arr[0]);
maxSum(arr, N);
return 0;
}
|
Java
import java.io.*;
class GFG{
static void maxSum( int [] arr, int N)
{
int ans = 0 ;
for ( int i = N - 1 ; i >= 0 ; i--)
{
int t = i;
if (t + arr[i] < N)
{
arr[i] += arr[t + arr[i]];
}
}
for ( int i = 0 ; i < N; i++)
{
System.out.print(arr[i] + " " );
}
}
public static void main(String[] args)
{
int [] arr = { 1 , 2 , 7 , 1 , 8 };
int N = arr.length;
maxSum(arr, N);
}
}
|
Python3
def maxSum(arr, N):
ans = 0 ;
for i in range (N - 1 , - 1 , - 1 ):
t = i;
if (t + arr[i] < N):
arr[i] + = arr[t + arr[i]];
for i in range (N):
print (arr[i], end = " " );
if __name__ = = '__main__' :
arr = [ 1 , 2 , 7 , 1 , 8 ];
N = len (arr);
maxSum(arr, N);
|
C#
using System;
class GFG {
static void maxSum( int [] arr, int N)
{
for ( int i = N - 1; i >= 0; i--)
{
int t = i;
if (t + arr[i] < N)
{
arr[i] += arr[t + arr[i]];
}
}
for ( int i = 0; i < N; i++)
{
Console.Write(arr[i] + " " );
}
}
static public void Main()
{
int [] arr = { 1, 2, 7, 1, 8 };
int N = arr.Length;
maxSum(arr, N);
}
}
|
Javascript
<script>
function maxSum(arr, N)
{
let ans = 0;
for (let i = N - 1; i >= 0; i--) {
let t = i;
if (t + arr[i] < N) {
arr[i] += arr[t + arr[i]];
}
}
for (let i = 0; i < N; i++) {
document.write(arr[i] + ' ' );
}
}
let arr = [ 1, 2, 7, 1, 8 ];
let N = arr.length;
maxSum(arr, N);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
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