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

• Difficulty Level : Basic
• Last Updated : 08 Sep, 2022

Given an array of size . 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 ``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, 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);` `    ``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, 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

 ` 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

 ``

Output

`2`

Complexity Analysis:

• Time Complexity: O(n)
• Auxiliary Space: O(1)

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