Maximum length sub-array which satisfies the given conditions
Last Updated :
07 Dec, 2022
Given a binary array arr[], the task is to find the length of the longest sub-array of the given array such that if the sub-array is divided into two equal-sized sub-arrays then both the sub-arrays either contain all 0s or all 1s. For example, the two sub-arrays must be of the form {0, 0, 0, 0} and {1, 1, 1, 1} or {1, 1, 1} and {0, 0, 0} and not {0, 0, 0} and {0, 0, 0}
Examples:
Input: arr[] = {1, 1, 1, 0, 0, 1, 1}
Output: 4
{1, 1, 0, 0} and {0, 0, 1, 1} are the maximum length valid sub-arrays.
Input: arr[] = {1, 1, 0, 0, 0, 1, 1, 1, 1}
Output: 6
{0, 0, 0, 1, 1, 1} is the only valid sub-array with maximum length.
Approach: For every two consecutive elements of the array say arr[i] and arr[j] where j = i + 1, treat them as the middle two elements of the required sub-array. In order for this sub-array to be a valid sub-array arr[i] must not be equal to arr[j]. If it can be a valid sub-array then its size is 2. Now, try to extend this sub-array to a bigger size by decrementing i and incrementing j at the same time and all the elements before index i and after index j must be equal to arr[i] and arr[j] respectively. Print the size of the longest such sub-array found so far.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the maximum length
// of the required sub-array
int maxLength(int arr[], int n)
{
int maxLen = 0;
// For the first consecutive
// pair of elements
int i = 0;
int j = i + 1;
// While a consecutive pair
// can be selected
while (j < n) {
// If current pair forms a
// valid sub-array
if (arr[i] != arr[j]) {
// 2 is the length of the
// current sub-array
maxLen = max(maxLen, 2);
// To extend the sub-array both ways
int l = i - 1;
int r = j + 1;
// While elements at indices l and r
// are part of a valid sub-array
while (l >= 0 && r < n && arr[l] == arr[i]
&& arr[r] == arr[j]) {
l--;
r++;
}
// Update the maximum length so far
maxLen = max(maxLen, 2 * (r - j));
}
// Select the next consecutive pair
i++;
j = i + 1;
}
// Return the maximum length
return maxLen;
}
// Driver code
int main()
{
int arr[] = { 1, 1, 1, 0, 0, 1, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << maxLength(arr, n);
return 0;
}
// Java implementation of the approach
class GFG {
// Function to return the maximum length
// of the required sub-array
static int maxLength(int arr[], int n)
{
int maxLen = 0;
// For the first consecutive
// pair of elements
int i = 0;
int j = i + 1;
// While a consecutive pair
// can be selected
while (j < n) {
// If current pair forms a
// valid sub-array
if (arr[i] != arr[j]) {
// 2 is the length of the
// current sub-array
maxLen = Math.max(maxLen, 2);
// To extend the sub-array both ways
int l = i - 1;
int r = j + 1;
// While elements at indices l and r
// are part of a valid sub-array
while (l >= 0 && r < n && arr[l] == arr[i] && arr[r] == arr[j]) {
l--;
r++;
}
// Update the maximum length so far
maxLen = Math.max(maxLen, 2 * (r - j));
}
// Select the next consecutive pair
i++;
j = i + 1;
}
// Return the maximum length
return maxLen;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 1, 1, 0, 0, 1, 1 };
int n = arr.length;
System.out.println(maxLength(arr, n));
}
}
// This code is contributed by AnkitRai01
# Python3 implementation of the approach
# Function to return the maximum length
# of the required sub-array
def maxLength(arr, n):
maxLen = 0
# For the first consecutive
# pair of elements
i = 0
j = i + 1
# While a consecutive pair
# can be selected
while (j < n):
# If current pair forms a
# valid sub-array
if (arr[i] != arr[j]):
# 2 is the length of the
# current sub-array
maxLen = max(maxLen, 2)
# To extend the sub-array both ways
l = i - 1
r = j + 1
# While elements at indices l and r
# are part of a valid sub-array
while (l >= 0 and r < n and arr[l] == arr[i]
and arr[r] == arr[j]):
l-= 1
r+= 1
# Update the maximum length so far
maxLen = max(maxLen, 2 * (r - j))
# Select the next consecutive pair
i+= 1
j = i + 1
# Return the maximum length
return maxLen
# Driver code
arr =[1, 1, 1, 0, 0, 1, 1]
n = len(arr)
print(maxLength(arr, n))
# This code is contributed by mohit kumar 29
// C# implementation of the approach
using System;
class GFG {
// Function to return the maximum length
// of the required sub-array
static int maxLength(int[] arr, int n)
{
int maxLen = 0;
// For the first consecutive
// pair of elements
int i = 0;
int j = i + 1;
// While a consecutive pair
// can be selected
while (j < n) {
// If current pair forms a
// valid sub-array
if (arr[i] != arr[j]) {
// 2 is the length of the
// current sub-array
maxLen = Math.Max(maxLen, 2);
// To extend the sub-array both ways
int l = i - 1;
int r = j + 1;
// While elements at indices l and r
// are part of a valid sub-array
while (l >= 0 && r < n && arr[l] == arr[i] && arr[r] == arr[j]) {
l--;
r++;
}
// Update the maximum length so far
maxLen = Math.Max(maxLen, 2 * (r - j));
}
// Select the next consecutive pair
i++;
j = i + 1;
}
// Return the maximum length
return maxLen;
}
// Driver code
public static void Main(String[] args)
{
int[] arr = { 1, 1, 1, 0, 0, 1, 1 };
int n = arr.Length;
Console.WriteLine(maxLength(arr, n));
}
}
// This code is contributed by 29AjayKumar
<script>
// JavaScript implementation of the approach
// Function to return the maximum length
// of the required sub-array
function maxLength(arr, n)
{
let maxLen = 0;
// For the first consecutive
// pair of elements
let i = 0;
let j = i + 1;
// While a consecutive pair
// can be selected
while (j < n) {
// If current pair forms a
// valid sub-array
if (arr[i] != arr[j]) {
// 2 is the length of the
// current sub-array
maxLen = Math.max(maxLen, 2);
// To extend the sub-array both ways
let l = i - 1;
let r = j + 1;
// While elements at indices l and r
// are part of a valid sub-array
while (l >= 0 && r < n && arr[l] == arr[i]
&& arr[r] == arr[j]) {
l--;
r++;
}
// Update the maximum length so far
maxLen = Math.max(maxLen, 2 * (r - j));
}
// Select the next consecutive pair
i++;
j = i + 1;
}
// Return the maximum length
return maxLen;
}
// Driver code
let arr = [ 1, 1, 1, 0, 0, 1, 1 ];
let n = arr.length;
document.write(maxLength(arr, n));
</script>
Time Complexity: O(n2), where n is the length of the given array.
Auxiliary Space: O(1)
Alternate approach: We can maintain the max length of previous similar elements and try to form subarray with the next different contiguous element and maximize the subarray length.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function to return the maximum length
// of the required sub-array
int maxLength(int a[], int n)
{
// To store the maximum length
// for a valid subarray
int maxLen = 0;
// To store the count of contiguous
// similar elements for previous
// group and the current group
int prev_cnt = 0, curr_cnt = 1;
for (int i = 1; i < n; i++) {
// If current element is equal to
// the previous element then it is
// a part of the same group
if (a[i] == a[i - 1])
curr_cnt++;
// Else update the previous group
// and start counting elements
// for the new group
else {
prev_cnt = curr_cnt;
curr_cnt = 1;
}
// Update the maximum possible length for a group
maxLen = max(maxLen, min(prev_cnt, curr_cnt));
}
// Return the maximum length of the valid subarray
return (2 * maxLen);
}
// Driver code
int main()
{
int arr[] = { 1, 1, 1, 0, 0, 1, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << maxLength(arr, n);
return 0;
}
// Java implementation of the approach
class GFG
{
// Function to return the maximum length
// of the required sub-array
static int maxLength(int a[], int n)
{
// To store the maximum length
// for a valid subarray
int maxLen = 0;
// To store the count of contiguous
// similar elements for previous
// group and the current group
int prev_cnt = 0, curr_cnt = 1;
for (int i = 1; i < n; i++)
{
// If current element is equal to
// the previous element then it is
// a part of the same group
if (a[i] == a[i - 1])
curr_cnt++;
// Else update the previous group
// and start counting elements
// for the new group
else
{
prev_cnt = curr_cnt;
curr_cnt = 1;
}
// Update the maximum possible length for a group
maxLen = Math.max(maxLen,
Math.min(prev_cnt, curr_cnt));
}
// Return the maximum length
// of the valid subarray
return (2 * maxLen);
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 1, 1, 0, 0, 1, 1 };
int n = arr.length;
System.out.println(maxLength(arr, n));
}
}
// This code is contributed by Rajput-Ji
# Python3 implementation of the approach
# Function to return the maximum length
# of the required sub-array
def maxLength(a, n):
# To store the maximum length
# for a valid subarray
maxLen = 0;
# To store the count of contiguous
# similar elements for previous
# group and the current group
prev_cnt = 0; curr_cnt = 1;
for i in range(1, n):
# If current element is equal to
# the previous element then it is
# a part of the same group
if (a[i] == a[i - 1]):
curr_cnt += 1;
# Else update the previous group
# and start counting elements
# for the new group
else:
prev_cnt = curr_cnt;
curr_cnt = 1;
# Update the maximum possible
# length for a group
maxLen = max(maxLen, min(prev_cnt,
curr_cnt));
# Return the maximum length
# of the valid subarray
return (2 * maxLen);
# Driver code
arr = [ 1, 1, 1, 0, 0, 1, 1 ];
n = len(arr);
print(maxLength(arr, n));
# This code is contributed by Rajput-Ji
// C# implementation of the approach
using System;
class GFG
{
// Function to return the maximum length
// of the required sub-array
static int maxLength(int[] a, int n)
{
// To store the maximum length
// for a valid subarray
int maxLen = 0;
// To store the count of contiguous
// similar elements for previous
// group and the current group
int prev_cnt = 0, curr_cnt = 1;
for (int i = 1; i < n; i++)
{
// If current element is equal to
// the previous element then it is
// a part of the same group
if (a[i] == a[i - 1])
curr_cnt++;
// Else update the previous group
// and start counting elements
// for the new group
else
{
prev_cnt = curr_cnt;
curr_cnt = 1;
}
// Update the maximum possible length for a group
maxLen = Math.Max(maxLen,
Math.Min(prev_cnt, curr_cnt));
}
// Return the maximum length
// of the valid subarray
return (2 * maxLen);
}
// Driver code
public static void Main()
{
int[] arr = { 1, 1, 1, 0, 0, 1, 1 };
int n = arr.Length;
Console.WriteLine(maxLength(arr, n));
}
}
// This code is contributed by Code_Mech.
<script>
// JavaScript implementation of the approach
// Function to return the maximum length
// of the required sub-array
function maxLength(a, n)
{
// To store the maximum length
// for a valid subarray
let maxLen = 0;
// To store the count of contiguous
// similar elements for previous
// group and the current group
let prev_cnt = 0, curr_cnt = 1;
for (let i = 1; i < n; i++) {
// If current element is equal to
// the previous element then it is
// a part of the same group
if (a[i] == a[i - 1])
curr_cnt++;
// Else update the previous group
// and start counting elements
// for the new group
else {
prev_cnt = curr_cnt;
curr_cnt = 1;
}
// Update the maximum possible length for a group
maxLen = Math.max(maxLen, Math.min(prev_cnt, curr_cnt));
}
// Return the maximum length of the valid subarray
return (2 * maxLen);
}
// Driver code
let arr = [ 1, 1, 1, 0, 0, 1, 1 ];
let n = arr.length;
document.write(maxLength(arr, n));
</script>
Time Complexity: O(n), where n is the length of the given array.
Auxiliary Space: O(1)
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