Data Modelling & AI Data Structure & Algorithm

Divide binary array into three equal parts with same value

31 July 2024

0

Given an array A of length n such that it contains only ‘0s’ and ‘1s’. The task is to divide the array into THREE different non-empty parts such that all of these parts represent the same binary value(in decimals).

If it is possible, return any [i, j] with i+1 < j, such that:

A[0], A[1], …, A[i] is the first part.
A[i+1], A[i+2], …, A[j-1] is the second part.
A[j], A[j+1], …, A[n- 1] is the third part. Note: All three parts should have equal binary values. However, If it is not possible, return [-1, -1].

Examples:

Input : A = [1, 1, 1, 1, 1, 1]
Output : [1, 4]
All three parts are,
A[0] to A[1] first part,
A[2] to A[3] second part,
A[4] to A[5] third part.

Input : A = [1, 0, 0, 1, 0, 1]
Output : [0, 4]

Naive Approach

The idea is to find every possible index pair i and j with i+1<j to divide the array into three parts. Then find the decimal value of each part and if the decimal value of all three parts are equal then return/print that i and j.

Steps to implement-

Find the size of the input array/vector
Declare a vector to store the final answer
Now run two nested for loops to choose index i and j with i+1<j such that we will get three parts of array
After that initialize three variables with a value of 0 to find the decimal value of those three parts
Run a for loop from i to 0 to find the decimal value of the first part
Run a for loop from j-1 to i+1 to find the decimal value of the second part
Run a for loop from N-1 to j to find the decimal value of the third part
When decimal value of all three part becomes equal then print or return i,j value
In last if nothing got printed or returned then print or return -1,-1

Code

C++

// C++ implementation of the 
// above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return required 
// interval answer.
vector<int> ThreeEqualParts(vector<int> arr)
{
    //Size of input vector
    int N=arr.size();
     
    //To store answer
   vector<int> ans;  
    
   //Check for different possible i,j
   for(int i=0;i<N-2;i++){
       for(int j=i+2;j<N;j++){
            
           //Initialize decimal value of all three part to zero
           int val1=0;
           int val2=0;
           int val3=0;
            
           int temp;
            
           //Finding decimal value of first part
           temp=1;
           for(int k=i;k>=0;k--){
             val1+=temp*arr[k];
             temp=temp*2;
           }
            
           //Finding decimal value of second part
           temp=1;
           for(int k=j-1;k>=i+1;k--){
             val2+=temp*arr[k];
             temp=temp*2;
           }
            
           //Finding decimal value of third part
           temp=1;
           for(int k=N-1;k>=j;k--){
             val3+=temp*arr[k];
             temp=temp*2;
           }
           
           //When all three part has equal value
           if(val1==val2 && val2==val3 && val1==val3){
               ans.push_back(i);
               ans.push_back(j);
               return ans;
           }
       }
   }
    
   //If no such value is possible then 
   //we need to return [-1,-1]
   ans.push_back(-1);
   ans.push_back(-1);
   return ans;
}
 
// Driver Code
int main()
{
    vector<int> arr = {1, 1, 1, 1, 1, 1}; 
 
    // Output required result
    vector<int> res = ThreeEqualParts(arr);
 
    for(auto it :res)
        cout << it << " ";
 
    return 0;
}

Java

import java.util.ArrayList;
import java.util.List;
 
public class Main {
 
    public static List<Integer> threeEqualParts(List<Integer> arr) {
        int N = arr.size();
        List<Integer> ans = new ArrayList<>();
 
        for (int i = 0; i < N - 2; i++) {
            for (int j = i + 2; j < N; j++) {
                int val1 = 0, val2 = 0, val3 = 0;
                int temp;
 
                for (int k = i; k >= 0; k--) {
                    temp = (int) Math.pow(2, i - k);
                    val1 += arr.get(k) * temp;
                }
 
                for (int k = j - 1; k >= i + 1; k--) {
                    temp = (int) Math.pow(2, j - k - 1);
                    val2 += arr.get(k) * temp;
                }
 
                for (int k = N - 1; k >= j; k--) {
                    temp = (int) Math.pow(2, N - 1 - k);
                    val3 += arr.get(k) * temp;
                }
 
                if (val1 == val2 && val2 == val3 && val1 == val3) {
                    ans.add(i);
                    ans.add(j);
                    return ans;
                }
            }
        }
 
        ans.add(-1);
        ans.add(-1);
        return ans;
    }
 
    public static void main(String[] args) {
        List<Integer> arr = List.of(1, 1, 1, 1, 1, 1);
        List<Integer> res = threeEqualParts(arr);
        System.out.println(res);
    }
}

Python3

from typing import List
 
def threeEqualParts(arr: List[int]) -> List[int]:
    N = len(arr)
    ans = []
 
    for i in range(N - 2):
        for j in range(i + 2, N):
            val1, val2, val3 = 0, 0, 0
            temp = 1
 
            for k in range(i, -1, -1):
                val1 += arr[k] * temp
                temp *= 2
 
            temp = 1
            for k in range(j - 1, i, -1):
                val2 += arr[k] * temp
                temp *= 2
 
            temp = 1
            for k in range(N - 1, j - 1, -1):
                val3 += arr[k] * temp
                temp *= 2
 
            if val1 == val2 == val3:
                ans.append(i)
                ans.append(j)
                return ans
 
    ans.append(-1)
    ans.append(-1)
    return ans
 
arr = [1, 1, 1, 1, 1, 1]
res = threeEqualParts(arr)
print(res)

C#

using System;
using System.Collections.Generic;
 
public class ThreeEqualParts {
    public static List<int> FindEqualParts(List<int> arr) {
        int N = arr.Count;
        List<int> ans = new List<int>();
        for(int i=0; i<N-2; i++) {
            for(int j=i+2; j<N; j++) {
                int val1=0, val2=0, val3=0;
                int temp;
                for(int k=i; k>=0; k--) {
                    val1 += arr[k] * (int)Math.Pow(2, i-k);
                }
                for(int k=j-1; k>=i+1; k--) {
                    val2 += arr[k] * (int)Math.Pow(2, j-1-k);
                }
                for(int k=N-1; k>=j; k--) {
                    val3 += arr[k] * (int)Math.Pow(2, N-1-k);
                }
                if(val1 == val2 && val2 == val3) {
                    ans.Add(i);
                    ans.Add(j);
                    return ans;
                }
            }
        }
        ans.Add(-1);
        ans.Add(-1);
        return ans;
    }
 
    public static void Main() {
        List<int> arr = new List<int> {1, 1, 1, 1, 1, 1};
        List<int> res = FindEqualParts(arr);
        foreach(int it in res) {
            Console.Write(it + " ");
        }
    }
}

PHP

<?php
    function threeEqualParts($arr) {
    // Size of input array
    $N = count($arr);
     
    // To store answer
    $ans = array();  
    
    // Check for different possible i, j
    for ($i = 0; $i < $N - 2; $i++) {
        for ($j = $i + 2; $j < $N; $j++) {
            
            // Initialize decimal value of all three part to zero
            $val1 = 0;
            $val2 = 0;
            $val3 = 0;
            
            $temp;
            
            // Finding decimal value of first part
            $temp = 1;
            for ($k = $i; $k >= 0; $k--) {
                $val1 += $temp * $arr[$k];
                $temp *= 2;
            }
            
            // Finding decimal value of second part
            $temp = 1;
            for ($k = $j - 1; $k >= $i + 1; $k--) {
                $val2 += $temp * $arr[$k];
                $temp *= 2;
            }
            
            // Finding decimal value of third part
            $temp = 1;
            for ($k = $N - 1; $k >= $j; $k--) {
                $val3 += $temp * $arr[$k];
                $temp *= 2;
            }
           
            // When all three part has equal value
            if ($val1 == $val2 && $val2 == $val3 && $val1 == $val3) {
                array_push($ans, $i);
                array_push($ans, $j);
                return $ans;
            }
        }
    }
    
    // If no such value is possible then 
    // we need to return [-1,-1]
    array_push($ans, -1);
    array_push($ans, -1);
    return $ans;
}
 
// Driver code
$arr = array(1, 1, 1, 1, 1, 1);
  
// Output required result
$res = threeEqualParts($arr);
  
foreach ($res as $it)
    echo $it . " ";
 
?>

Javascript

function findEqualParts(arr) {
    const N = arr.length;
    const ans = [];
    for(let i=0; i<N-2; i++) {
        for(let j=i+2; j<N; j++) {
            let val1=0, val2=0, val3=0;
            let temp;
            for(let k=i; k>=0; k--) {
                val1 += arr[k] * Math.pow(2, i-k);
            }
            for(let k=j-1; k>=i+1; k--) {
                val2 += arr[k] * Math.pow(2, j-1-k);
            }
            for(let k=N-1; k>=j; k--) {
                val3 += arr[k] * Math.pow(2, N-1-k);
            }
            if(val1 === val2 && val2 === val3) {
                ans.push(i);
                ans.push(j);
                return ans;
            }
        }
    }
    ans.push(-1);
    ans.push(-1);
    return ans;
}
 
const arr = [1, 1, 1, 1, 1, 1];
const res = findEqualParts(arr);
console.log(res);

Output

1 4

Time Complexity: O(N³), because of two nested loops to find i and j, and then loops to find the decimal value of all three parts
Auxiliary Space: O(1), because no extra space has been used

Another Approach:

Lets say total number of ones in A be S. Since every part has the same number of ones, thus all parts should have K = S / 3 ones. If S isn’t divisible by 3 i:e S % 3 != 0, then the task is impossible. Now we will find the position of the 1st, K-th, K+1-th, 2K-th, 2K+1-th, and 3K-th one. The positions of these ones will form 3 intervals: [i1, j1], [i2, j2], [i3, j3]. (If there are only 3 ones, then the intervals are each length 1.) Between the intervals, there may be some number of zeros.

The zeros after third interval j3 must be included in each part: say there are z of them (z = length of (S) – j3). So the first part, [i1, j1], is now [i1, j1+z]. Similarly, the second part, [i2, j2], is now [i2, j2+z]. If all this is actually possible, then the final answer is [j1+z, j2+z+1].

Below is the implementation of above approach.

C++

// C++ implementation of the 
// above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return required 
// interval answer.
vector<int> ThreeEqualParts(vector<int> A)
{
    int imp[] = {-1, -1};
    vector<int> IMP(imp, imp + 2);
 
    // Finding total number of ones
    int Sum = accumulate(A.begin(), 
                         A.end(), 0);
 
    if (Sum % 3)
    {
        return IMP;
    }
 
    int K = Sum / 3;
 
    // Array contains all zeros.
    if (K == 0)
    {
        return {0, (int)A.size() - 1};
    }
 
    vector<int> interval;
    int S = 0;
 
    for (int i = 0 ;i < A.size(); i++)
    {
        int x = A[i];
        if (x)
        {
            S += x;
            if (S == 1 or S == K + 1 or S == 2 * K + 1)
            {
                interval.push_back(i);
            }
            if (S == K or S == 2 * K or S == 3 * K)
            {
                interval.push_back(i);
            }
        }
    }
 
    int i1 = interval[0], j1 = interval[1],
        i2 = interval[2], j2 = interval[3],
        i3 = interval[4], j3 = interval[5];
 
    vector<int> a(A.begin() + i1, A.begin() + j1 + 1);
    vector<int> b(A.begin() + i2, A.begin() + j2 + 1);
    vector<int> c(A.begin() + i3, A.begin() + j3 + 1);
 
    // The array is in the form 
    // W [i1, j1] X [i2, j2] Y [i3, j3] Z
    // where [i1, j1] is a block of 1s, and so on.
    if (!((a == b) and (b == c)))
    {
        return {-1, -1};
    }
 
    // x, y, z: the number of zeros
    // after part 1, 2, 3
    int x = i2 - j1 - 1;
    int y = i3 - j2 - 1;
    int z = A.size() - j3 - 1;
 
    if (x < z or y < z)
    {
        return IMP;
    } 
 
    // appending extra zeros at end of 
    // first and second interval
    j1 += z;
    j2 += z;
    return {j1, j2 + 1}; 
}
 
// Driver Code
int main()
{
    vector<int> A = {1, 1, 1, 1, 1, 1}; 
 
    // Output required result
    vector<int> res = ThreeEqualParts(A);
 
    for(auto it :res)
        cout << it << " ";
 
    return 0;
}
 
// This code is contributed 
// by Harshit Saini 

Java

// Java implementation of the 
// above approach
import java.util.*;
 
class GFG
{
 
// Function to return required 
// interval answer.
public static int[] ThreeEqualParts(int[] A)
{
    int IMP[] = new int[]{-1, -1};
 
    // Finding total number of ones
    int Sum = Arrays.stream(A).sum();
     
    if ((Sum % 3) != 0)
    {
        return IMP;
    }
 
    int K = Sum / 3;
 
    // Array contains all zeros.
    if (K == 0)
    {
        return new int[]{0, A.length - 1};
    }
 
    ArrayList<Integer> interval = 
                   new ArrayList<Integer>();
    int S = 0;
 
    for (int i = 0 ;i < A.length; i++)
    {
        int x = A[i];
        if (x != 0)
        {
            S += x;
            if ((S == 1) || (S == K + 1) || 
                (S == 2 * K + 1))
            {
                interval.add(i);
            }
            if ((S == K) || (S == 2 * K) || 
                (S == 3 * K))
            {
                interval.add(i);
            }
        }
    }
 
    int i1 = interval.get(0), j1 = interval.get(1),
        i2 = interval.get(2), j2 = interval.get(3),
        i3 = interval.get(4), j3 = interval.get(5);
 
    int [] a = Arrays.copyOfRange(A, i1, j1 + 1);
    int [] b = Arrays.copyOfRange(A, i2, j2 + 1);
    int [] c = Arrays.copyOfRange(A, i3, j3 + 1);
 
    // The array is in the form
    // W [i1, j1] X [i2, j2] Y [i3, j3] Z
    // where [i1, j1] is a block of 1s, and so on.
    if (!(Arrays.equals(a, b) && 
          Arrays.equals(b, c)))
    {
        return new int[]{-1, -1};
    }
 
    // x, y, z: 
    // the number of zeros after part 1, 2, 3
    int x = i2 - j1 - 1;
    int y = i3 - j2 - 1;
    int z = A.length - j3 - 1;
 
    if (x < z || y < z)
    {
        return IMP;
    } 
 
    // appending extra zeros at end 
    // of first and second interval
    j1 += z;
    j2 += z;
    return new int[]{j1, j2 + 1}; 
}
 
// Driver Code
public static void main(String []args)
{
    int[] A = new int[]{1, 1, 1, 1, 1, 1}; 
 
    // Output required result
    int[] res = ThreeEqualParts(A);
 
    System.out.println(Arrays.toString(res));
}
}
 
// This code is contributed 
// by Harshit Saini 

Python3

# Python implementation of the above approach
 
# Function to return required interval answer.
def ThreeEqualParts(A):
    IMP = [-1, -1]
 
    # Finding total number of ones
    Sum = sum(A)
 
    if Sum % 3:
        return IMP
 
    K = Sum / 3
 
    # Array contains all zeros.
    if K == 0:
        return [0, len(A) - 1]
 
    interval = []
    S = 0
 
    for i, x in enumerate(A):
        if x:
            S += x
            if S in {1, K + 1, 2 * K + 1}:
                interval.append(i)
            if S in {K, 2 * K, 3 * K}:
                interval.append(i)
 
    i1, j1, i2, j2, i3, j3 = interval
 
    # The array is in the form W [i1, j1] X [i2, j2] Y [i3, j3] Z
    # where [i1, j1] is a block of 1s, and so on.
    if not(A[i1:j1 + 1] == A[i2:j2 + 1] == A[i3:j3 + 1]):
        return [-1, -1]
 
    # x, y, z: the number of zeros after part 1, 2, 3
    x = i2 - j1 - 1
    y = i3 - j2 - 1
    z = len(A) - j3 - 1
 
    if x < z or y < z:
        return IMP
 
    # appending extra zeros at end of first and second interval
    j1 += z
    j2 += z
    return [j1, j2 + 1]
 
 
# Driver Program
A = [1, 1, 1, 1, 1, 1]
 
# Output required result
print(ThreeEqualParts(A))
 
# This code is written by
# Sanjit_Prasad

C#

// C# implementation of the 
// above approach
using System;
using System.Linq;
using System.Collections.Generic;
 
class GFG
{
 
// Function to return required
// interval answer.
static int[] ThreeEqualParts(int[] A)
{
 
    int []IMP = new int[]{-1, -1};
 
    // Finding total number of ones
    int Sum = A.Sum();
     
    if ((Sum % 3) != 0)
    {
        return IMP;
    }
 
 
    int K = Sum / 3;
 
    // Array contains all zeros.
    if (K == 0)
    {
        return new int[]{0, A.Length - 1};
    }
 
    List<int> interval = new List<int>();
 
    int S = 0;
 
    for (int i = 0 ;i < A.Length; i++)
    {
        int x = A[i];
        if (x != 0)
        {
            S += x;
            if ((S == 1) || (S == K + 1) || 
                (S == 2 * K + 1))
            {
                interval.Add(i);
            }
            if ((S == K) || (S == 2 * K) || 
                (S == 3 * K))
            {
                interval.Add(i);
            }
        }
    }
 
    int i1 = interval[0], j1 = interval[1],
        i2 = interval[2], j2 = interval[3],
        i3 = interval[4], j3 = interval[5];
 
    var a = A.Skip(i1).Take(j1 - i1 + 1).ToArray();
    var b = A.Skip(i2).Take(j2 - i2 + 1).ToArray();
    var c = A.Skip(i3).Take(j3 - i3 + 1).ToArray();
 
    // The array is in the form 
    // W [i1, j1] X [i2, j2] Y [i3, j3] Z
    // where [i1, j1] is a block of 1s,
    // and so on.
    if (!(Enumerable.SequenceEqual(a,b) && 
          Enumerable.SequenceEqual(b,c)))
    {
        return new int[]{-1, -1};
    }
 
    // x, y, z: the number of zeros
    // after part 1, 2, 3
    int X = i2 - j1 - 1;
    int y = i3 - j2 - 1;
    int z = A.Length - j3 - 1;
 
    if (X < z || y < z)
    {
        return IMP;
    } 
 
    // appending extra zeros at end 
    // of first and second interval
    j1 += z;
    j2 += z;
    return new int[]{j1, j2 + 1}; 
}
 
// Driver Code
public static void Main()
{
    int[] A = new int[]{1, 1, 1, 1, 1, 1}; 
 
    // Output required result
    int[] res = ThreeEqualParts(A);
 
    Console.WriteLine(string.Join(" ", res));
}
}
 
// This code is contributed 
// by Harshit Saini 

PHP

<?php
// PHP implementation of the
// above approach
 
// Function to return required 
// interval answer.
function ThreeEqualParts($A)
{
 
    $IMP = array(-1, -1);
 
    // Finding total number of ones
    $Sum = array_sum($A);
 
    if ($Sum % 3)
    {
        return $IMP;
    }
    $K = $Sum / 3;
 
    // Array contains all zeros.
    if ($K == 0)
    {
        return array(0, count(A, 
                     COUNT_NORMAL) - 1);
    }
 
    $interval = array();
    $S = 0;
 
    for ($i = 0; 
         $i < count($A, COUNT_NORMAL); $i++)
    {
        $x = $A[$i];
        if ($x)
        {
            $S += $x;
            if ($S == 1 or $S == $K + 1 or
                $S == 2 * $K + 1)
            {
                array_push($interval,$i);
            }
            if ($S == $K or $S == 2 * $K or
                $S == 3 * $K)
            {
                array_push($interval, $i);
            }
        }
    }
 
    $i1 = $interval[0]; $j1 = $interval[1];
    $i2 = $interval[2]; $j2 = $interval[3];
    $i3 = $interval[4]; $j3 = $interval[5];
 
    $a = array_slice($A, $i1, $j1 - $i1 + 1);
    $b = array_slice($A, $i1, $j2 - $i2 + 1);
    $c = array_slice($A, $i1, $j3 - $i3 + 1);
 
    // The array is in the form  
    // W [i1, j1] X [i2, j2] Y [i3, j3] Z
    // where [i1, j1] is a block of 1s,
    // and so on.
    if (!($a == $b and $b == $c))
    {
        return array(-1, -1);
    }
 
    // x, y, z: the number of zeros
    // after part 1, 2, 3
    $x = $i2 - $j1 - 1;
    $y = $i3 - $j2 - 1;
    $z = count($A) - $j3 - 1;
 
    if ($x < $z or $y < $z)
    {
        return $IMP;
    } 
 
    // appending extra zeros at end 
    // of first and second interval
    $j1 += $z;
    $j2 += $z;
    return array($j1, $j2 + 1); 
}
 
// Driver Code
$A = array(1, 1, 1, 1, 1, 1); 
 
// Output required result
$res = ThreeEqualParts($A);
 
foreach ($res as $key => $value) 
{
    echo $value." ";
}
 
// This code is contributed 
// by Harshit Saini 
?>

Javascript

//JavaScript implementation of the above approach
 
//function to compare two arrays
  function isEqual(a, b)
  {
    // If length is not equal
    if(a.length!=b.length)
     return false;
    else
    {
      
    // Comparing each element of array
     for(var i=0;i<a.length;i++)
     if(a[i]!=b[i])
      return false;
      return true;
    }
  }
 
// Function to return required interval answer.
function ThreeEqualParts(A)
{
    var IMP = [-1, -1];
 
    // Finding total number of ones
    var Sum = A.reduce(function (x, y) {
        return x + y;
    }, 0);
 
    if (Sum % 3 > 0)
        return IMP;
 
    var K = Sum / 3;
 
    // Array contains all zeros.
    if (K == 0)
        return [0, A.length - 1];
 
    var interval = [];
    var S = 0;
     
    for (var i = 0; i < A.length; i++)
    {
        x = A[i];
        if (x)
        {
            S += x;
            if (S == 1 || S == K + 1 || S == 2 * K + 1)
                interval.push(i);
            if (S == K || S == 2 * K || S == 3 * K)
                interval.push(i);
        }
    }
    var i1 = interval[0];
    var j1 = interval[1];
    var i2 = interval[2];
    var j2 = interval[3];
    var i3 = interval[4];
    var j3 = interval[5];
 
 
    // The array is in the form W [i1, j1] X [i2, j2] Y [i3, j3] Z
    // where [i1, j1] is a block of 1s, and so on.
    if (!(isEqual(A.slice(i1, j1 + 1), A.slice(i2, j2 + 1)) || isEqual(A.slice(i2, j2 + 1), A.slice(i3,j3 + 1))))
        return [-1, -1];
 
    // x, y, z: the number of zeros after part 1, 2, 3
    var x = i2 - j1 - 1;
    var y = i3 - j2 - 1;
    var z = A.length - j3 - 1;
 
    if (x < z || y < z)
        return IMP;
 
    // appending extra zeros at end of first and second interval
    j1 += z;
    j2 += z;
    return [j1, j2 + 1];
}
 
 
// Driver Program
var A = [1, 1, 1, 1, 1, 1];
 
// Output required result
console.log(ThreeEqualParts(A));
 
// This code is written by phasing17

Output

1 4

Complexity Analysis:

Time Complexity: O(N), where N is the length of S.
Space Complexity: O(N)

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Divide binary array into three equal parts with same value

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