Split array into three equal sum segments
Last Updated :
29 Oct, 2024
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Given an integer array arr[], the task is to divide the array into three non-empty contiguous segments with equal sum. In other words, we need to return an index pair [i, j], such that sum(arr[0…i]) = sum(arr[i+1…j]) = sum(arr[j+1…n-1]).
Note: If it is impossible to divide the array into three non-empty contiguous segments, return [-1, -1].
Examples:
Input: arr[] = [1, 3, 4, 0, 4]
Output: [1, 2]
Explanation: 3 equal sum segments are: arr[0…1], arr[2…2] and arr[3…4] each having sum = 4.Input: arr[] = [2, 3, 4]
Output: [-1, -1]
Explanation: No three segments exist which has equal sum.Input: arr[] = [1, -1, 1, -1, 1, -1, 1, -1]
Output: [1, 3]
Explanation: 3 equal sum segments are: arr[0…1], arr[2…3] and arr[4…7] each having sum = 0.
Table of Content
[Naive Approach] By finding all possible partitions – O(n^3) Time and O(1) Space
The idea is to try all possible partitions using two variables, say i and j such that the first segment will be arr[0…i], the second segment will be arr[i+1…j] and the third segment will be arr[j+1…n-1]. For any partition, if the sum of all three segments are equal, then we can return [i, j] as a valid partition. Otherwise, return [-1, -1].
// C++ program to find if the array can be divided into
// three segments by finding all possible partitions
#include <iostream>
#include <vector>
using namespace std;
// function to find the sum of subarray arr[start...end]
int findSum(vector<int> &arr, int start, int end) {
int sum = 0;
for(int i = start; i <= end; i++)
sum += arr[i];
return sum;
}
// function to return the index pair of equal sum segments
vector<int> findSplit(vector<int> &arr) {
int n = arr.size();
// First partition
for(int i = 0; i < n - 2; i++) {
// Second Partition
for(int j = i + 1; j < n - 1; j++) {
// Find sum of all three segments
int sum1 = findSum(arr, 0, i);
int sum2 = findSum(arr, i + 1, j);
int sum3 = findSum(arr, j + 1, n - 1);
// If all three segments have equal sum,
// then return true
if(sum1 == sum2 && sum2 == sum3)
return {i, j};
}
}
// No possible index pair found
return {-1, -1};
}
int main() {
vector<int> arr = {1, 3, 4, 0, 4};
vector<int> res = findSplit(arr);
cout << res[0] << " " << res[1];
return 0;
}
// C program to find if the array can be divided into
// three segments by finding all possible partitions
#include <stdio.h>
// function to find the sum of subarray arr[start...end]
int findSum(int *arr, int start, int end) {
int sum = 0;
for(int i = start; i <= end; i++)
sum += arr[i];
return sum;
}
// function to return the index pair of equal sum segments
int* findSplit(int *arr, int n) {
int *res = (int *)malloc(2 * sizeof(int));
res[0] = -1;
res[1] = -1;
// First partition
for(int i = 0; i < n - 2; i++) {
// Second Partition
for(int j = i + 1; j < n - 1; j++) {
// Find sum of all three segments
int sum1 = findSum(arr, 0, i);
int sum2 = findSum(arr, i + 1, j);
int sum3 = findSum(arr, j + 1, n - 1);
// If all three segments have equal sum,
// then return true
if(sum1 == sum2 && sum2 == sum3) {
res[0] = i;
res[1] = j;
return res;
}
}
}
// No possible index pair found
return res;
}
int main() {
int arr[] = {1, 3, 4, 0, 4};
int n = sizeof(arr) / sizeof(arr[0]);
int *res = findSplit(arr, n);
printf("%d %d", res[0], res[1]);
return 0;
}
// Java program to find if the array can be divided into
// three segments by finding all possible partitions
class GfG {
// function to find the sum of subarray arr[start...end]
static int findSum(int[] arr, int start, int end) {
int sum = 0;
for (int i = start; i <= end; i++)
sum += arr[i];
return sum;
}
// function to return the index pair of equal sum segments
static int[] findSplit(int[] arr) {
int n = arr.length;
// First partition
for (int i = 0; i < n - 2; i++) {
// Second Partition
for (int j = i + 1; j < n - 1; j++) {
// Find sum of all three segments
int sum1 = findSum(arr, 0, i);
int sum2 = findSum(arr, i + 1, j);
int sum3 = findSum(arr, j + 1, n - 1);
// If all three segments have equal sum,
// then return true
if (sum1 == sum2 && sum2 == sum3)
return new int[]{i, j};
}
}
// No possible index pair found
return new int[]{-1, -1};
}
public static void main(String[] args) {
int[] arr = {1, 3, 4, 0, 4};
int[] res = findSplit(arr);
System.out.println(res[0] + " " + res[1]);
}
}
# Python program to find if the array can be divided into
# three segments by finding all possible partitions
# function to find the sum of subarray arr[start...end]
def findSum(arr, start, end):
sum_ = 0
for i in range(start, end + 1):
sum_ += arr[i]
return sum_
# function to return the index pair of equal sum segments
def findSplit(arr):
n = len(arr)
# First partition
for i in range(n - 2):
# Second Partition
for j in range(i + 1, n - 1):
# Find sum of all three segments
sum1 = findSum(arr, 0, i)
sum2 = findSum(arr, i + 1, j)
sum3 = findSum(arr, j + 1, n - 1)
# If all three segments have equal sum,
# then return true
if sum1 == sum2 and sum2 == sum3:
return [i, j]
# No possible index pair found
return [-1, -1]
if __name__ == "__main__":
arr = [1, 3, 4, 0, 4]
res = findSplit(arr)
print(res[0], res[1])
// C# program to find if the array can be divided into
// three segments by finding all possible partitions
using System;
using System.Collections.Generic;
class GfG {
// function to find the sum of subarray arr[start...end]
static int findSum(int[] arr, int start, int end) {
int sum = 0;
for (int i = start; i <= end; i++)
sum += arr[i];
return sum;
}
// function to return the index pair of equal sum segments
static int[] findSplit(int[] arr) {
int n = arr.Length;
// First partition
for (int i = 0; i < n - 2; i++) {
// Second Partition
for (int j = i + 1; j < n - 1; j++) {
// Find sum of all three segments
int sum1 = findSum(arr, 0, i);
int sum2 = findSum(arr, i + 1, j);
int sum3 = findSum(arr, j + 1, n - 1);
// If all three segments have equal sum,
// then return true
if (sum1 == sum2 && sum2 == sum3)
return new int[] { i, j };
}
}
// No possible index pair found
return new int[] { -1, -1 };
}
static void Main() {
int[] arr = { 1, 3, 4, 0, 4 };
int[] res = findSplit(arr);
Console.WriteLine(res[0] + " " + res[1]);
}
}
// JavaScript program to find if the array can be divided
// into three segments by finding all possible partitions
// function to find the sum of subarray arr[start...end]
function findSum(arr, start, end) {
let sum = 0;
for (let i = start; i <= end; i++)
sum += arr[i];
return sum;
}
// function to return the index pair of equal sum segments
function findSplit(arr) {
let n = arr.length;
// First partition
for (let i = 0; i < n - 2; i++) {
// Second Partition
for (let j = i + 1; j < n - 1; j++) {
// Find sum of all three segments
let sum1 = findSum(arr, 0, i);
let sum2 = findSum(arr, i + 1, j);
let sum3 = findSum(arr, j + 1, n - 1);
// If all three segments have equal sum,
// then return true
if (sum1 === sum2 && sum2 === sum3)
return [i, j];
}
}
// No possible index pair found
return [-1, -1];
}
// Driver Code
const arr = [1, 3, 4, 0, 4];
const res = findSplit(arr);
console.log(res[0] + " " + res[1]);
Output
1 2
[Expected Approach] Finding first two segments- O(n) Time and O(1) Space
To split the array into three equal segments, we first need to make sure that the total sum of the array is divisible by 3. Then, as we iterate through the array, we calculate the running sum. If running sum becomes equal to one-third of the total, we reset the running sum to zero and store the index as the first element of the index pair. If we find another index for which the running sum becomes equal to one-third of the total and there are still elements left for a third segment, then store the index as the second element of the index pair and return the index pair.
We only need the first two segments with sum equal to one-third of the total because the remaining subarray will always be the third segment.
Working:
// C++ program to find if the array can be divided into
// three segments by finding first two segments
#include <iostream>
#include <vector>
using namespace std;
// function to return the index pair of equal sum segments
vector<int> findSplit(vector<int> &arr) {
vector<int> res;
int total = 0;
for (int ele : arr)
total += ele;
// If the total sum is not divisible by 3,
// it's impossible to split the array
if (total % 3 != 0) {
res = {-1, -1};
return res;
}
// Keep track of the sum of current segment
int currSum = 0;
for (int i = 0; i < arr.size(); i++) {
currSum += arr[i];
// If the valid segment is found, store its index
// and reset current sum to zero
if (currSum == total / 3) {
currSum = 0;
res.push_back(i);
// If two valid segments are found and third non
// empty segment is possible, return the index pair
if (res.size() == 2 && i < arr.size() - 1)
return res;
}
}
// If no index pair is possible
res = {-1, -1};
return res;
}
int main() {
vector<int> arr = {1, 3, 4, 0, 4};
vector<int> res = findSplit(arr);
cout << res[0] << " " << res[1];
return 0;
}
// C program to find if the array can be divided into
// three segments by finding first two segments
#include <stdio.h>
int *findSplit(int *arr, int n) {
int *res = (int *)malloc(2 * sizeof(int));
int resIdx = 0;
int total = 0;
for (int i = 0; i < n; i++)
total += arr[i];
// If the total sum is not divisible by 3,
// it's impossible to split the array
if (total % 3 != 0) {
res[0] = -1;
res[1] = -1;
return res;
}
// Keep track of the sum of current segment
int currSum = 0;
for (int i = 0; i < n; i++) {
currSum += arr[i];
// If the valid segment is found, store its index
// and reset current sum to zero
if (currSum == total / 3) {
currSum = 0;
res[resIdx++] = i;
// If two valid segments are found and third non
// empty segment is possible, return true
if (resIdx == 2 && i < n - 1) {
return res;
}
}
}
res[0] = -1;
res[1] = -1;
return res;
}
int main()
{
int arr[] = {1, 3, 4, 0, 4};
int size = sizeof(arr) / sizeof(arr[0]);
int *res = findSplit(arr, size);
printf("%d %d\n", res[0], res[1]);
return 0;
}
// Java program to find if the array can be divided into
// three segments by finding first two segments
import java.util.Arrays;
class GfG {
static int[] findSplit(int[] arr) {
int[] res = new int[2];
int n = arr.length;
int resIdx = 0;
int total = 0;
for (int i = 0; i < n; i++)
total += arr[i];
// If the total sum is not divisible by 3,
// it's impossible to split the array
if (total % 3 != 0) {
res[0] = -1;
res[1] = -1;
return res;
}
// Keep track of the sum of current segment
int currSum = 0;
for (int i = 0; i < n; i++) {
currSum += arr[i];
// If the valid segment is found, store its index
// and reset current sum to zero
if (currSum == total / 3) {
currSum = 0;
res[resIdx++] = i;
// If two valid segments are found and third non
// empty segment is possible, return true
if (resIdx == 2 && i < n - 1) {
return res;
}
}
}
res[0] = -1;
res[1] = -1;
return res;
}
public static void main(String[] args) {
int[] arr = {1, 3, 4, 0, 4};
int[] res = findSplit(arr);
System.out.println(res[0] + " " + res[1]);
}
}
# Python program to find if the array can be divided into
# three segments by finding first two segments
# function to return the index pair of equal sum segments
def findSplit(arr):
res = []
total = 0
for ele in arr:
total += ele
# If the total sum is not divisible by 3,
# it's impossible to split the array
if total % 3 != 0:
res = [-1, -1]
return res
# Keep track of the sum of current segment
currSum = 0
for i in range(len(arr)):
currSum += arr[i]
# If the valid segment is found, store its index
# and reset current sum to zero
if currSum == total / 3:
currSum = 0
res.append(i)
# If two valid segments are found and third non
# empty segment is possible, return the index pair
if len(res) == 2 and i < len(arr) - 1:
return res
# If no index pair is possible
res = [-1, -1]
return res
if __name__ == "__main__":
arr = [1, 3, 4, 0, 4]
res = findSplit(arr)
print(res[0], res[1])
// C# program to find if the array can be divided into
// three segments by finding first two segments
using System;
class GfG {
static int[] findSplit(int[] arr) {
int[] res = new int[2];
int n = arr.Length;
int resIdx = 0;
int total = 0;
for (int i = 0; i < n; i++)
total += arr[i];
// If the total sum is not divisible by 3,
// it's impossible to split the array
if (total % 3 != 0) {
res[0] = -1;
res[1] = -1;
return res;
}
// Keep track of the sum of current segment
int currSum = 0;
for (int i = 0; i < n; i++) {
currSum += arr[i];
// If the valid segment is found, store its index
// and reset current sum to zero
if (currSum == total / 3) {
currSum = 0;
res[resIdx++] = i;
// If two valid segments are found and third non
// empty segment is possible, return true
if (resIdx == 2 && i < n - 1) {
return res;
}
}
}
res[0] = -1;
res[1] = -1;
return res;
}
static void Main() {
int[] arr = { 1, 3, 4, 0, 4 };
int[] res = findSplit(arr);
Console.WriteLine($"{res[0]} {res[1]}");
}
}
// JavaScript program to find if the array can be divided into
// three segments by finding first two segments
// function to return the index pair of equal sum segments
function findSplit(arr) {
let res = [];
let total = 0;
for (let ele of arr)
total += ele;
// If the total sum is not divisible by 3,
// it's impossible to split the array
if (total % 3 !== 0) {
res = [-1, -1];
return res;
}
// Keep track of the sum of current segment
let currSum = 0;
for (let i = 0; i < arr.length; i++) {
currSum += arr[i];
// If the valid segment is found, store its index
// and reset current sum to zero
if (currSum === total / 3) {
currSum = 0;
res.push(i);
// If two valid segments are found and third non
// empty segment is possible, return the index pair
if (res.length === 2 && i < arr.length - 1)
return res;
}
}
// If no index pair is possible
res = [-1, -1];
return res;
}
// Driver Code
const arr = [1, 3, 4, 0, 4];
const res = findSplit(arr);
console.log(res[0], res[1]);
Output
1 2
