Partition the array into three equal sum segments
Last Updated :
06 Jul, 2021
Given an array of n integers, we have to partition the array into three segments such that all the segments have an equal sum. Segment sum is the sum of all the elements in the segment.
Examples:
Input : 1, 3, 6, 2, 7, 1, 2, 8
Output : [1, 3, 6], [2, 7, 1], [2, 8]
Input : 7, 6, 1, 7
Output : [7], [6, 1], [7]
Input : 7, 6, 2, 7
Output : Cannot divide the array into segments
A simple solution is to consider all pairs of indexes and, for every pair, check if it divides the array into three equal parts. If yes, then return true. The time complexity of this solution is O(n2)
An efficient approach is to use two auxiliary arrays and store the prefix and suffix array sum in these arrays respectively. We then use the two-pointer approach, with variable 'i' pointing to the start of the prefix array and variable 'j' pointing to the end of the suffix array. If pre[i] > suf[j], then decrement 'j', otherwise increment 'i'.
We maintain a variable whose value is the total sum of the array and whenever we encounter pre[i] = total_sum / 3 or suf[j] = total_sum / 3, we store the value of i or j respectively as segment boundaries.
C++
// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
// First segment's end index
static int pos1 = -1;
// Third segment's start index
static int pos2 = -1;
// This function returns true if the array
// can be divided into three equal sum segments
bool equiSumUtil(int arr[],int n)
{
// Prefix Sum Array
int pre[n];
int sum = 0;
for (int i = 0; i < n; i++)
{
sum += arr[i];
pre[i] = sum;
}
// Suffix Sum Array
int suf[n];
sum = 0;
for (int i = n - 1; i >= 0; i--)
{
sum += arr[i];
suf[i] = sum;
}
// Stores the total sum of the array
int total_sum = sum;
int i = 0, j = n - 1;
while (i < j - 1)
{
if (pre[i] == total_sum / 3)
{
pos1 = i;
}
if (suf[j] == total_sum / 3)
{
pos2 = j;
}
if (pos1 != -1 && pos2 != -1)
{
// We can also take pre[pos2 - 1] - pre[pos1] ==
// total_sum / 3 here.
if (suf[pos1 + 1] - suf[pos2] == total_sum / 3)
{
return true;
}
else
{
return false;
}
}
if (pre[i] < suf[j])
{
i++;
}
else
{
j--;
}
}
return false;
}
void equiSum(int arr[],int n)
{
bool ans = equiSumUtil(arr,n);
if (ans)
{
cout << "First Segment : ";
for (int i = 0; i <= pos1; i++)
{
cout << arr[i] << " ";
}
cout << endl;
cout << "Second Segment : ";
for (int i = pos1 + 1; i < pos2; i++)
{
cout << arr[i] << " ";
}
cout << endl;
cout << "Third Segment : ";
for (int i = pos2; i < n; i++)
{
cout << arr[i] << " ";
}
cout<<endl;
}
else
{
cout << "Array cannot be divided into three equal sum segments";
}
}
// Driver code
int main()
{
int arr[] = { 1, 3, 6, 2, 7, 1, 2, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
equiSum(arr,n);
return 0;
}
// This code is contributed by mits
public class Main {
// First segment's end index
public static int pos1 = -1;
// Third segment's start index
public static int pos2 = -1;
// This function returns true if the array
// can be divided into three equal sum segments
public static boolean equiSumUtil(int[] arr)
{
int n = arr.length;
// Prefix Sum Array
int[] pre = new int[n];
int sum = 0;
for (int i = 0; i < n; i++) {
sum += arr[i];
pre[i] = sum;
}
// Suffix Sum Array
int[] suf = new int[n];
sum = 0;
for (int i = n - 1; i >= 0; i--) {
sum += arr[i];
suf[i] = sum;
}
// Stores the total sum of the array
int total_sum = sum;
int i = 0, j = n - 1;
while (i < j - 1) {
if (pre[i] == total_sum / 3) {
pos1 = i;
}
if (suf[j] == total_sum / 3) {
pos2 = j;
}
if (pos1 != -1 && pos2 != -1) {
// We can also take pre[pos2 - 1] - pre[pos1] ==
// total_sum / 3 here.
if (suf[pos1 + 1] - suf[pos2] == total_sum / 3) {
return true;
}
else {
return false;
}
}
if (pre[i] < suf[j]) {
i++;
}
else {
j--;
}
}
return false;
}
public static void equiSum(int[] arr)
{
boolean ans = equiSumUtil(arr);
if (ans) {
System.out.print("First Segment : ");
for (int i = 0; i <= pos1; i++) {
System.out.print(arr[i] + " ");
}
System.out.println();
System.out.print("Second Segment : ");
for (int i = pos1 + 1; i < pos2; i++) {
System.out.print(arr[i] + " ");
}
System.out.println();
System.out.print("Third Segment : ");
for (int i = pos2; i < arr.length; i++) {
System.out.print(arr[i] + " ");
}
System.out.println();
}
else {
System.out.println("Array cannot be " +
"divided into three equal sum segments");
}
}
public static void main(String[] args)
{
int[] arr = { 1, 3, 6, 2, 7, 1, 2, 8 };
equiSum(arr);
}
}
# Python3 implementation of the given approach
# This function returns true if the array
# can be divided into three equal sum segments
def equiSumUtil(arr, pos1, pos2):
n = len(arr);
# Prefix Sum Array
pre = [0] * n;
sum = 0;
for i in range(n):
sum += arr[i];
pre[i] = sum;
# Suffix Sum Array
suf = [0] * n;
sum = 0;
for i in range(n - 1, -1, -1):
sum += arr[i];
suf[i] = sum;
# Stores the total sum of the array
total_sum = sum;
i = 0;
j = n - 1;
while (i < j - 1):
if (pre[i] == total_sum // 3):
pos1 = i;
if (suf[j] == total_sum // 3):
pos2 = j;
if (pos1 != -1 and pos2 != -1):
# We can also take pre[pos2 - 1] - pre[pos1] ==
# total_sum / 3 here.
if (suf[pos1 + 1] -
suf[pos2] == total_sum // 3):
return [True, pos1, pos2];
else:
return [False, pos1, pos2];
if (pre[i] < suf[j]):
i += 1;
else:
j -= 1;
return [False, pos1, pos2];
def equiSum(arr):
pos1 = -1;
pos2 = -1;
ans = equiSumUtil(arr, pos1, pos2);
pos1 = ans[1];
pos2 = ans[2];
if (ans[0]):
print("First Segment : ", end = "");
for i in range(pos1 + 1):
print(arr[i], end = " ");
print("");
print("Second Segment : ", end = "");
for i in range(pos1 + 1, pos2):
print(arr[i], end = " ");
print("");
print("Third Segment : ", end = "");
for i in range(pos2, len(arr)):
print(arr[i], end = " ");
print("");
else:
println("Array cannot be divided into",
"three equal sum segments");
# Driver Code
arr = [1, 3, 6, 2, 7, 1, 2, 8 ];
equiSum(arr);
# This code is contributed by mits
// C# implementation of the approach
using System;
class GFG
{
// First segment's end index
public static int pos1 = -1;
// Third segment's start index
public static int pos2 = -1;
// This function returns true if the array
// can be divided into three equal sum segments
public static bool equiSumUtil(int[] arr)
{
int n = arr.Length;
// Prefix Sum Array
int[] pre = new int[n];
int sum = 0,i;
for (i = 0; i < n; i++)
{
sum += arr[i];
pre[i] = sum;
}
// Suffix Sum Array
int[] suf = new int[n];
sum = 0;
for (i = n - 1; i >= 0; i--)
{
sum += arr[i];
suf[i] = sum;
}
// Stores the total sum of the array
int total_sum = sum;
int j = n - 1;
i = 0;
while (i < j - 1)
{
if (pre[i] == total_sum / 3)
{
pos1 = i;
}
if (suf[j] == total_sum / 3)
{
pos2 = j;
}
if (pos1 != -1 && pos2 != -1)
{
// We can also take pre[pos2 - 1] - pre[pos1] ==
// total_sum / 3 here.
if (suf[pos1 + 1] - suf[pos2] == total_sum / 3)
{
return true;
}
else
{
return false;
}
}
if (pre[i] < suf[j])
{
i++;
}
else
{
j--;
}
}
return false;
}
public static void equiSum(int[] arr)
{
bool ans = equiSumUtil(arr);
if (ans)
{
Console.Write("First Segment : ");
for (int i = 0; i <= pos1; i++)
{
Console.Write(arr[i] + " ");
}
Console.WriteLine();
Console.Write("Second Segment : ");
for (int i = pos1 + 1; i < pos2; i++)
{
Console.Write(arr[i] + " ");
}
Console.WriteLine();
Console.Write("Third Segment : ");
for (int i = pos2; i < arr.Length; i++)
{
Console.Write(arr[i] + " ");
}
Console.WriteLine();
}
else
{
Console.WriteLine("Array cannot be " +
"divided into three equal sum segments");
}
}
public static void Main(String[] args)
{
int[] arr = { 1, 3, 6, 2, 7, 1, 2, 8 };
equiSum(arr);
}
}
// This code contributed by Rajput-Ji
<?php
// PHP implementation of the given approach
// First segment's end index
$pos1 = -1;
// Third segment's start index
$pos2 = -1;
// This function returns true if the array
// can be divided into three equal sum segments
function equiSumUtil($arr)
{
global $pos2, $pos1;
$n = count($arr);
// Prefix Sum Array
$pre = array_fill(0, $n, 0);
$sum = 0;
for ($i = 0; $i < $n; $i++)
{
$sum += $arr[$i];
$pre[$i] = $sum;
}
// Suffix Sum Array
$suf = array_fill(0, $n, 0);
$sum = 0;
for ($i = $n - 1; $i >= 0; $i--)
{
$sum += $arr[$i];
$suf[$i] = $sum;
}
// Stores the total sum of the array
$total_sum = $sum;
$i = 0;
$j = $n - 1;
while ($i < $j - 1)
{
if ($pre[$i] == $total_sum / 3)
{
$pos1 = $i;
}
if ($suf[$j] == $total_sum / 3)
{
$pos2 = $j;
}
if ($pos1 != -1 && $pos2 != -1)
{
// We can also take pre[pos2 - 1] - pre[pos1] ==
// total_sum / 3 here.
if ($suf[$pos1 + 1] -
$suf[$pos2] == $total_sum / 3)
{
return true;
}
else
{
return false;
}
}
if ($pre[$i] < $suf[$j])
{
$i++;
}
else
{
$j--;
}
}
return false;
}
function equiSum($arr)
{
global $pos2,$pos1;
$ans = equiSumUtil($arr);
if ($ans)
{
print("First Segment : ");
for ($i = 0; $i <= $pos1; $i++)
{
print($arr[$i] . " ");
}
print("\n");
print("Second Segment : ");
for ($i = $pos1 + 1; $i < $pos2; $i++)
{
print($arr[$i] . " ");
}
print("\n");
print("Third Segment : ");
for ($i = $pos2; $i < count($arr); $i++)
{
print($arr[$i] . " ");
}
print("\n");
}
else
{
println("Array cannot be divided into ",
"three equal sum segments");
}
}
// Driver Code
$arr = array(1, 3, 6, 2, 7, 1, 2, 8 );
equiSum($arr);
// This code is contributed by mits
?>
<script>
// C# implementation of the approach
// First segment's end index
let pos1 = -1;
// Third segment's start index
let pos2 = -1;
// This function returns true if the array
// can be divided into three equal sum segments
function equiSumUtil(arr)
{
let n = arr.length;
// Prefix Sum Array
let pre = new Array(n);
let sum = 0,i;
for (i = 0; i < n; i++)
{
sum += arr[i];
pre[i] = sum;
}
// Suffix Sum Array
let suf = new Array(n);
sum = 0;
for (i = n - 1; i >= 0; i--)
{
sum += arr[i];
suf[i] = sum;
}
// Stores the total sum of the array
let total_sum = sum;
let j = n - 1;
i = 0;
while (i < j - 1)
{
if (pre[i] == total_sum / 3)
{
pos1 = i;
}
if (suf[j] == total_sum / 3)
{
pos2 = j;
}
if (pos1 != -1 && pos2 != -1)
{
// We can also take pre[pos2 - 1] - pre[pos1] ==
// total_sum / 3 here.
if (suf[pos1 + 1] - suf[pos2] == total_sum / 3)
{
return true;
}
else
{
return false;
}
}
if (pre[i] < suf[j])
{
i++;
}
else
{
j--;
}
}
return false;
}
function equiSum(arr)
{
let ans = equiSumUtil(arr);
if (ans)
{
document.write("First Segment : ");
for (let i = 0; i <= pos1; i++)
{
document.write(arr[i] + " ");
}
document.write("<br>");
document.write("Second Segment : ");
for (let i = pos1 + 1; i < pos2; i++)
{
document.write(arr[i] + " ");
}
document.write("<br>");
document.write("Third Segment : ");
for (let i = pos2; i < arr.length; i++)
{
document.write(arr[i] + " ");
}
document.write("<br>");
}
else
{
document.writeLine("Array cannot be" +
" divided into three equal sum segments");
}
}
let arr =[1, 3, 6, 2, 7, 1, 2, 8];
equiSum(arr);
</script>
Output: First Segment : 1 3 6
Second Segment : 2 7 1
Third Segment : 2 8
Time Complexity : O(n)
Auxiliary Space : O(n)
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