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Find the index of the left pointer after possible moves in the array
  • Difficulty Level : Basic
  • Last Updated : 05 Apr, 2021

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:
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:



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++




// 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;
}

Java




// 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..

Python3




# Python3 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

C#




// 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..

PHP




<?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..
?>

Javascript




<script>
 
// Javascript 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
    let 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
let a = [ 2, 7, 9, 8, 7 ];
let n = a.length;
 
document.write(getIndex(a, n));
 
// This code is contributed by Mayank Tyagi
     
</script>
Output: 
2

 

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