Find maximum equal sum of every three stacks

Given three stacks of the positive numbers, the task is to find the possible equal maximum sum of the stacks with the removal of top elements allowed. Stacks are represented as an array, and the first index of the array represent the top element of the stack.

Examples: 

Input : stack1[] = { 3, 10}
  stack2[] = { 4, 5 }
  stack3[] = { 2, 1 }
Output : 0
Sum can only be equal after removing all elements 
from all stacks.

The idea is to compare the sum of each stack, if they are not same, remove the top element of the stack having the maximum sum.

Algorithm for solving this problem: 

1. Find the sum of all elements of in individual stacks.
2. If the sum of all three stacks is the same, then this is the maximum sum.
3. Else remove the top element of the stack having the maximum sum among three of stacks. Repeat step 1 and step 2.

The approach works because elements are positive. To make sum equal, we must remove some element from stack having more sum, and we can only remove from the top.

Below is the implementation of this approach: 

C++

// C++ program to calculate maximum sum with equal
// stack sum.
#include <bits/stdc++.h>
using namespace std;

// Returns maximum possible equal sum of three stacks
// with removal of top elements allowed
int maxSum(int stack1[], int stack2[], int stack3[], int n1,
           int n2, int n3)
{
    int sum1 = 0, sum2 = 0, sum3 = 0;

    // Finding the initial sum of stack1.
    for (int i = 0; i < n1; i++)
        sum1 += stack1[i];

    // Finding the initial sum of stack2.
    for (int i = 0; i < n2; i++)
        sum2 += stack2[i];

    // Finding the initial sum of stack3.
    for (int i = 0; i < n3; i++)
        sum3 += stack3[i];

    // As given in question, first element is top
    // of stack..
    int top1 = 0, top2 = 0, top3 = 0;
    while (1) {
        // If any stack is empty
        if (top1 == n1 || top2 == n2 || top3 == n3)
            return 0;

        // If sum of all three stack are equal.
        if (sum1 == sum2 && sum2 == sum3)
            return sum1;

        // Finding the stack with maximum sum and
        // removing its top element.
        if (sum1 >= sum2 && sum1 >= sum3)
            sum1 -= stack1[top1++];
        else if (sum2 >= sum1 && sum2 >= sum3)
            sum2 -= stack2[top2++];
        else if (sum3 >= sum2 && sum3 >= sum1)
            sum3 -= stack3[top3++];
    }
}

// Driven Program
int main()
{
    int stack1[] = { 3, 2, 1, 1, 1 };
    int stack2[] = { 4, 3, 2 };
    int stack3[] = { 1, 1, 4, 1 };

    int n1 = sizeof(stack1) / sizeof(stack1[0]);
    int n2 = sizeof(stack2) / sizeof(stack2[0]);
    int n3 = sizeof(stack3) / sizeof(stack3[0]);

    cout << maxSum(stack1, stack2, stack3, n1, n2, n3)
         << endl;
    return 0;
}

Java

// JAVA Code for Find maximum sum possible
// equal sum of three stacks
class GFG {

    // Returns maximum possible equal sum of three
    // stacks with removal of top elements allowed
    public static int maxSum(int stack1[], int stack2[],
                            int stack3[], int n1, int n2,
                                               int n3)
    {
      int sum1 = 0, sum2 = 0, sum3 = 0;

      // Finding the initial sum of stack1.
      for (int i=0; i < n1; i++)
          sum1 += stack1[i];

      // Finding the initial sum of stack2.
      for (int i=0; i < n2; i++)
          sum2 += stack2[i];

      // Finding the initial sum of stack3.
      for (int i=0; i < n3; i++)
          sum3 += stack3[i];

      // As given in question, first element is top
      // of stack..
      int top1 =0, top2 = 0, top3 = 0;
      int ans = 0;
      while (true)
      {
          // If any stack is empty
          if (top1 == n1 || top2 == n2 || top3 == n3)
             return 0;

          // If sum of all three stack are equal.
          if (sum1 == sum2 && sum2 == sum3)
             return sum1;

          // Finding the stack with maximum sum and
          // removing its top element.
          if (sum1 >= sum2 && sum1 >= sum3)
             sum1 -= stack1[top1++];
          else if (sum2 >= sum1 && sum2 >= sum3)
             sum2 -= stack2[top2++];
          else if (sum3 >= sum2 && sum3 >= sum1)
             sum3 -= stack3[top3++];
       }
    }

    /* Driver program to test above function */
    public static void main(String[] args)
    {
          int stack1[] = { 3, 2, 1, 1, 1 };
          int stack2[] = { 4, 3, 2 };
          int stack3[] = { 1, 1, 4, 1 };

          int n1 = stack1.length;
          int n2 = stack2.length;
          int n3 = stack3.length;

          System.out.println(maxSum(stack1, stack2,
                               stack3, n1, n2, n3));
    }
  }
// This code is contributed by Arnav Kr. Mandal.

Python3

# Python program to calculate maximum sum with equal
# stack sum.
# Returns maximum possible equal sum of three stacks
# with removal of top elements allowed
def maxSum(stack1, stack2, stack3, n1, n2, n3):
    sum1, sum2, sum3 = 0, 0, 0

  # Finding the initial sum of stack1.
    for i in range(n1):
        sum1 += stack1[i]

  # Finding the initial sum of stack2.
    for i in range(n2):
        sum2 += stack2[i]

  # Finding the initial sum of stack3.
    for i in range(n3):
        sum3 += stack3[i]

  # As given in question, first element is top
  # of stack..
    top1, top2, top3 = 0, 0, 0
    ans = 0
    while (1):
      # If any stack is empty
        if (top1 == n1 or top2 == n2 or top3 == n3):
            return 0

      # If sum of all three stack are equal.
        if (sum1 == sum2 and sum2 == sum3):
            return sum1

      # Finding the stack with maximum sum and
      # removing its top element.
        if (sum1 >= sum2 and sum1 >= sum3):
            sum1 -= stack1[top1]
            top1=top1+1
        else if (sum2 >= sum1 and sum2 >= sum3):
            sum2 -= stack2[top2]
            top2=top2+1
        else if (sum3 >= sum2 and sum3 >= sum1):
            sum3 -= stack3[top3]
            top3=top3+1

# Driven Program
stack1 = [ 3, 2, 1, 1, 1 ]
stack2 = [ 4, 3, 2 ]
stack3 = [ 1, 1, 4, 1 ]

n1 = len(stack1)
n2 = len(stack2)
n3 = len(stack3)

print (maxSum(stack1, stack2, stack3, n1, n2, n3))

#This code is contributed by Afzal Ansari

C#

// C# Code for Find maximum sum with
// equal sum of three stacks
using System;

class GFG {

    // Returns maximum possible equal
    // sum of three stacks with removal
    // of top elements allowed
    public static int maxSum(int[] stack1,
               int[] stack2, int[] stack3,
                   int n1, int n2, int n3)
    {

        int sum1 = 0, sum2 = 0, sum3 = 0;

        // Finding the initial sum of
        // stack1.
        for (int i = 0; i < n1; i++)
            sum1 += stack1[i];

        // Finding the initial sum of
        // stack2.
        for (int i = 0; i < n2; i++)
            sum2 += stack2[i];

        // Finding the initial sum of
        // stack3.
        for (int i = 0; i < n3; i++)
            sum3 += stack3[i];

        // As given in question, first
        // element is top of stack..
        int top1 = 0, top2 = 0, top3 = 0;

        while (true) {

            // If any stack is empty
            if (top1 == n1 || top2 == n2
                            || top3 == n3)
                return 0;

            // If sum of all three stack
            // are equal.
            if (sum1 == sum2 && sum2 == sum3)
                return sum1;

            // Finding the stack with maximum
            // sum and removing its top element.
            if (sum1 >= sum2 && sum1 >= sum3)
                sum1 -= stack1[top1++];
            else if (sum2 >= sum1 && sum2 >= sum3)
                sum2 -= stack2[top2++];
            else if (sum3 >= sum2 && sum3 >= sum1)
                sum3 -= stack3[top3++];
        }
    }

    /* Driver program to test above function */
    public static void Main()
    {
        int[] stack1 = { 3, 2, 1, 1, 1 };
        int[] stack2 = { 4, 3, 2 };
        int[] stack3 = { 1, 1, 4, 1 };

        int n1 = stack1.Length;
        int n2 = stack2.Length;
        int n3 = stack3.Length;

        Console.Write(maxSum(stack1, stack2,
                        stack3, n1, n2, n3));
    }
}

// This code is contributed by nitin mittal.

PHP

<?php
// PHP program to calculate maximum
// sum with equal stack sum.

// Returns maximum possible
// equal sum of three stacks
// with removal of top elements
// allowed
function maxSum($stack1, $stack2, $stack3,
                            $n1, $n2, $n3)
{
    $sum1 = 0; $sum2 = 0; $sum3 = 0;

    // Finding the initial sum of stack1.
    for ($i = 0; $i < $n1; $i++)
        $sum1 += $stack1[$i];

    // Finding the initial sum of stack2.
    for ($i = 0; $i < $n2; $i++)
        $sum2 += $stack2[$i];

    // Finding the initial sum of stack3.
    for ($i = 0; $i < $n3; $i++)
        $sum3 += $stack3[$i];

    // As given in question,
    // first element is top
    // of stack..
    $top1 =0;
    $top2 = 0;
    $top3 = 0;
    $ans = 0;
    while (1)
    {

        // If any stack is empty
        if ($top1 == $n1 || $top2 == $n2 ||
                              $top3 == $n3)
            return 0;

        // If sum of all three stack are equal.
        if ($sum1 == $sum2 && $sum2 == $sum3)
            return $sum1;

        // Finding the stack with
        // maximum sum and
        // removing its top element.
        if ($sum1 >= $sum2 && $sum1 >= $sum3)
                $sum1 -= $stack1[$top1++];

        else if ($sum2 >= $sum1 && $sum2 >=$sum3)
                $sum2 -= $stack2[$top2++];

        else if ($sum3 >= $sum2 && $sum3 >= $sum1)
                $sum3 -= $stack3[$top3++];
    }
}

// Driver Code
$stack1 = array(3, 2, 1, 1, 1);
$stack2 = array(4, 3, 2);
$stack3 = array(1, 1, 4, 1);

$n1 = sizeof($stack1);
$n2 = sizeof($stack2);
$n3 = sizeof($stack3);
echo maxSum($stack1, $stack2,
            $stack3, $n1,
            $n2, $n3) ;

// This code is contributed by nitin mittal
?>

Javascript

<script>

// JavaScript program to calculate maximum
// sum with equal stack sum.

// Returns maximum possible equal sum of three
// stacks with removal of top elements allowed
function maxSum(stack1, stack2,
                stack3, n1, n2, n3)
{
    let sum1 = 0, sum2 = 0, sum3 = 0;

    // Finding the initial sum of stack1.
    for(let i = 0; i < n1; i++)
        sum1 += stack1[i];

    // Finding the initial sum of stack2.
    for(let i = 0; i < n2; i++)
        sum2 += stack2[i];

    // Finding the initial sum of stack3.
    for(let i = 0; i < n3; i++)
        sum3 += stack3[i];

    // As given in question, first element
    // is top of stack.
    let top1 = 0, top2 = 0, top3 = 0;
    let ans = 0;

    while (true)
    {

        // If any stack is empty
        if (top1 == n1 || top2 == n2 ||
            top3 == n3)
            return 0;

        // If sum of all three stack are equal.
        if (sum1 == sum2 && sum2 == sum3)
            return sum1;

        // Finding the stack with maximum sum and
        // removing its top element.
        if (sum1 >= sum2 && sum1 >= sum3)
            sum1 -= stack1[top1++];
        else if (sum2 >= sum1 && sum2 >= sum3)
            sum2 -= stack2[top2++];
        else if (sum3 >= sum2 && sum3 >= sum1)
            sum3 -= stack3[top3++];
    }
}

// Driver Code
let stack1 = [ 3, 2, 1, 1, 1 ];
let stack2 = [ 4, 3, 2 ];
let stack3 = [ 1, 1, 4, 1 ];

let n1 = stack1.length;
let n2 = stack2.length;
let n3 = stack3.length;

document.write(maxSum(stack1, stack2,
                      stack3, n1, n2, n3));

// This code is contributed by souravghosh0416

</script>

Output

5

Time Complexity : O(n1 + n2 + n3) where n1, n2 and n3 are sizes of three stacks.

Auxiliary space: O(1) because using constant space for variables

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