Find the index of the left pointer after possible moves in the array

Given an array of size N. Move two pointers, one from the left side and one from the right side of the array, a pointer will only move forward if the sum of all the numbers it has already gone through is less than the sum of the numbers the other pointer has gone through. Continue the process while left pointer is less than the right pointer or no move is possible. Print the position of the left pointer in the end.

Note: 0-based indexing is considered for the array.

Examples:

Input: arr[] = {2, 7, 9, 8, 7}
Output: 2
Initial position : ptrL = 0, ptrR = 4 with sum 2 and 7 respectively
Move 1 : ptrL = 1, ptrR = 4 with sum 9 and 7
Move 2 : ptrL = 1, ptrR = 3 with sum 9 and 15
Move 3 : ptrL = 2, ptrR = 3 with sum 18 and 7

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



Approach: An efficient approach is to move from the left and from the right at the same time and maintaining the running sum for both the pointers.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find the index of the left pointer
  
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns the index of the left pointer
int getIndex(int a[], int n)
{
    // there's only one element in the array
    if(n == 1)
        return 0;
  
    // initially both are at end
    int ptrL = 0, ptrR = n-1, sumL = a[0], sumR = a[n-1];
  
    while (ptrR - ptrL > 1) {
        if (sumL < sumR) {
            ptrL++;
            sumL += a[ptrL];
        }
        else if (sumL > sumR) {
            ptrR--;
            sumR += a[ptrR];
        }
        else {
            break;
        }
    }
    return ptrL;
}
  
// Driver code
int main()
{
    int a[] = { 2, 7, 9, 8, 7 };
  
    int n = sizeof(a) / sizeof(a[0]);
  
    cout << getIndex(a, n);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find the index of the left pointer
  
import java.io.*;
  
class GFG {
  
  
// Function that returns the index of the left pointer
static int getIndex(int a[], int n)
{
    // there's only one element in the array
    if(n == 1)
        return 0;
  
    // initially both are at end
    int ptrL = 0, ptrR = n-1, sumL = a[0], sumR = a[n-1];
  
    while (ptrR - ptrL > 1) {
        if (sumL < sumR) {
            ptrL++;
            sumL += a[ptrL];
        }
        else if (sumL > sumR) {
            ptrR--;
            sumR += a[ptrR];
        }
        else {
            break;
        }
    }
    return ptrL;
}
  
// Driver code
  
  
    public static void main (String[] args) {
        int a[] = { 2, 7, 9, 8, 7 };
  
    int n =a.length;
  
    System.out.println ( getIndex(a, n));
    }
}
// This code is contributed by  anuj_67..

chevron_right


Python 3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 3 program to find the 
# index of the left pointer
  
# Function that returns the 
# index of the left pointer
def getIndex(a, n):
      
    # there's only one element 
    # in the array
    if(n == 1):
        return 0
  
    # initially both are at end
    ptrL = 0
    ptrR = n-1
    sumL = a[0]
    sumR = a[n-1]
  
    while (ptrR - ptrL > 1) :
        if (sumL < sumR) :
            ptrL += 1
            sumL += a[ptrL]
          
        elif (sumL > sumR) :
            ptrR -= 1
            sumR += a[ptrR]
          
        else :
            break
    return ptrL
  
# Driver code
if __name__ == "__main__":
      
    a = [ 2, 7, 9, 8, 7 ]
  
    n = len(a)
  
    print(getIndex(a, n))
      
# This code is contributed by
# ChitraNayal

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

//C# program to find the index of the left pointer
  
using System;
  
class GFG {
  
  
// Function that returns the index of the left pointer
static int getIndex(int []a, int n)
{
    // there's only one element in the array
    if(n == 1)
        return 0;
  
    // initially both are at end
    int ptrL = 0, ptrR = n-1, sumL = a[0], sumR = a[n-1];
  
    while (ptrR - ptrL > 1) {
        if (sumL < sumR) {
            ptrL++;
            sumL += a[ptrL];
        }
        else if (sumL > sumR) {
            ptrR--;
            sumR += a[ptrR];
        }
        else {
            break;
        }
    }
    return ptrL;
}
  
// Driver code
  
  
    public static void Main () {
        int []a = { 2, 7, 9, 8, 7 };
  
    int n =a.Length;
  
    Console.WriteLine( getIndex(a, n));
    }
}
// This code is contributed by anuj_67..

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to find the index 
// of the left pointer
  
// Function that returns the index
// of the left pointer
function getIndex($a, $n)
{
    // there's only one element
    // in the array
    if($n == 1)
        return 0;
  
    // initially both are at end
    $ptrL = 0; $ptrR = $n - 1; 
    $sumL = $a[0]; $sumR = $a[$n - 1];
  
    while ($ptrR - $ptrL > 1)
    {
        if ($sumL < $sumR
        {
            $ptrL++;
            $sumL += $a[$ptrL];
        }
        else if ($sumL > $sumR
        {
            $ptrR--;
            $sumR += $a[$ptrR];
        }
        else 
        {
            break;
        }
    }
    return $ptrL;
}
  
// Driver code
$a = array( 2, 7, 9, 8, 7 );
  
$n = count($a);
  
echo getIndex($a, $n);
  
// This code is contributed by anuj_67..
?>

chevron_right


Output:

2


My Personal Notes arrow_drop_up

pawanasipugmailcom

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : vt_m, ChitraNayal