Given an array containing only 0s and 1s. For each index ‘i‘(0 index), find length of the longest alternating sub-array starting from ‘i‘ to ‘j‘ i.e., ai..j for i <= j < n. Alternating sub-array means that any two adjacent elements should be different.
Example:
Input: arr[] = {1, 0, 1, 0, 0, 1}
Output: 4 3 2 1 2 1
Explanation
Length for index '0': {1 0 1 0} => 4
Length for index '1': {0 1 0} => 3
Length for index '2': {1 0} => 2
Length for index '3': {0} => 1
Length for index '4': {0 1} => 2
Length for index '5': {1} => 1
Simple approach is to iterate for every index by using two ‘for loops’ for finding the length of each index. The outer loop picks a starting point ‘i’ and the inner loop considers all sub-arrays starting from ‘i’. Time complexity of this approach is O(n2) which is not sufficient for large value of ‘n’.
Better approach is to use Dynamic programming. First of all, let’s see how to check for alternating sub-array. To check whether two elements are alternating or not, we can simply take XOR(Ex-OR) of both of them and then compare with ‘0’ and ‘1’.
- If there XOR is ‘0’, that means both numbers are not alternating(equal elements).
- If there XOR is ‘1’, that means both are alternating(different elements)
Let len[i] denotes length of an alternating sub-array starting at position ‘i’. If arr[i] and arr[i+1] have different values, then len[i] will be one more than len[i+1]. Otherwise if they are same elements, len[i] will be just 1.
Below is the implementation of above approach:
C++
// C++ program to calculate longest alternating // sub-array for each index elements #include <bits/stdc++.h> using namespace std; // Function to calculate alternating sub- // array for each index of array elements void alternateSubarray(bool arr[], int n) { int len[n]; // Initialize the base state of len[] len[n - 1] = 1; // Calculating value for each element for (int i = n - 2; i >= 0; --i) { // If both elements are different // then add 1 to next len[i+1] if (arr[i] ^ arr[i + 1] == 1) len[i] = len[i + 1] + 1; // else initialize to 1 else len[i] = 1; } // Print lengths of binary subarrays. for (int i = 0; i < n; ++i) cout << len[i] << " "; } // Driver program int main() { bool arr[] = { 1, 0, 1, 0, 0, 1 }; int n = sizeof(arr) / sizeof(arr[0]); alternateSubarray(arr, n); return 0; } |
Java
// Java program to calculate longest alternating // sub-array for each index elements import java.util.*; import java.io.*; class GFG { // Function to calculate alternating sub- // array for each index of array elements static void alternateSubarray(boolean arr[], int n) { int len[] = new int[n]; // Initialize the base state of len[] len[n - 1] = 1; // Calculating value for each element for (int i = n - 2; i >= 0; --i) { // If both elements are different // then add 1 to next len[i+1] if (arr[i] ^ arr[i + 1] == true) len[i] = len[i + 1] + 1; // else initialize to 1 else len[i] = 1; } // Print lengths of binary subarrays. for (int i = 0; i < n; ++i) System.out.print(len[i] + " "); } // Driver code public static void main(String[] args) { boolean arr[] = {true, false, true, false, false, true}; int n = arr.length; alternateSubarray(arr, n); } } // This code is contributed by Anant Agarwal. |
Python3
# Python program to calculate # longest alternating # sub-array for each index elements # Function to calculate alternating sub- # array for each index of array elements def alternateSubarray(arr,n): len=[] for i in range(n+1): len.append(0) # Initialize the base state of len[] len[n - 1] = 1 # Calculating value for each element for i in range(n - 2,-1,-1): # If both elements are different # then add 1 to next len[i+1] if (arr[i] ^ arr[i + 1] == True): len[i] = len[i + 1] + 1 # else initialize to 1 else: len[i] = 1 # Print lengths of binary subarrays. for i in range(n): print(len[i] , " ",end="") # Driver code arr = [ True,False, True, False, False, True ] n = len(arr) alternateSubarray(arr, n) # This code is contributed # by Anant Agarwal. |
C#
// C# program to calculate longest // alternating sub-array for each // index elements using System; class GFG { // Function to calculate alternating sub- // array for each index of array elements static void alternateSubarray(bool []arr, int n) { int []len = new int[n]; // Initialize the base state of len[] len[n - 1] = 1; // Calculating value for each element for (int i = n - 2; i >= 0; --i) { // If both elements are different // then add 1 to next len[i+1] if (arr[i] ^ arr[i + 1] == true) len[i] = len[i + 1] + 1; // else initialize to 1 else len[i] = 1; } // Print lengths of binary subarrays. for (int i = 0; i < n; ++i) Console.Write(len[i] + " "); } // Driver code public static void Main() { bool []arr = {true, false, true, false, false, true}; int n = arr.Length; alternateSubarray(arr, n); } } // This code is contributed by vt_m. |
PHP
<?php // PHP program to calculate longest alternating // sub-array for each index elements // Function to calculate alternating sub- // array for each index of array elements function alternateSubarray(&$arr, $n) { $len = array_fill(0, $n, NULL); // Initialize the base state of len[] $len[$n - 1] = 1; // Calculating value for each element for ($i = $n - 2; $i >= 0; --$i) { // If both elements are different // then add 1 to next len[i+1] if ($arr[$i] ^ $arr[$i + 1] == 1) $len[$i] = $len[$i + 1] + 1; // else initialize to 1 else $len[$i] = 1; } // Print lengths of binary subarrays. for ($i = 0; $i < $n; ++$i) echo $len[$i] . " "; } // Driver Code $arr = array(1, 0, 1, 0, 0, 1); $n = sizeof($arr); alternateSubarray($arr, $n); // This code is contributed by ita_c ?> |
Javascript
<script> // JavaScript program to calculate longest alternating // sub-array for each index elements // Function to calculate alternating sub- // array for each index of array elements function alternateSubarray(arr, n) { let len = new Array(n); // Initialize the base state of len[] len[n - 1] = 1; // Calculating value for each element for (let i = n - 2; i >= 0; --i) { // If both elements are different // then add 1 to next len[i+1] if (arr[i] ^ arr[i + 1] == 1) len[i] = len[i + 1] + 1; // else initialize to 1 else len[i] = 1; } // Print lengths of binary subarrays. for (let i = 0; i < n; ++i) document.write(len[i] + " "); } // Driver program let arr = [ 1, 0, 1, 0, 0, 1 ]; let n = arr.length; alternateSubarray(arr, n); // This code is contributed by Surbhi Tyagi. </script> |
Output
4 3 2 1 2 1
Time complexity: O(n)
Auxiliary space: O(n)
Another Efficient approach is, instead of storing all the sub-array elements in len[] array, we can directly print it until mismatch(equal adjacent elements) is found. When a mismatch is found, we print count from current value to 0.
Implementation:
C++
// C++ program to calculate longest alternating // sub-array for each index elements #include <bits/stdc++.h> using namespace std; // Function to calculate alternating sub-array // for each index of array elements void alternateSubarray(bool arr[], int n) { // Initialize count variable for storing // length of sub-array int count = 1; // Initialize 'prev' variable which // indicates the previous element // while traversing for index 'i' int prev = arr[0]; for (int i = 1; i < n; ++i) { // If both elements are same // print elements because alternate // element is not found for current // index if ((arr[i] ^ prev) == 0) { // print count and decrement it. while (count) cout << count-- << " "; } // Increment count for next element ++count; // Re-initialize previous variable prev = arr[i]; } // If elements are still available after // traversing whole array, print it while (count) cout << count-- << " "; } // Driver program int main() { bool arr[] = { 1, 0, 1, 0, 0, 1 }; int n = sizeof(arr) / sizeof(arr[0]); alternateSubarray(arr, n); return 0; } |
Java
// Java program to calculate longest alternating // sub-array for each index elements import java.io.*; import java.util.*; class GFG { // Function to calculate alternating sub-array // for each index of array elements static void alternateSubarray(boolean arr[], int n) { // Initialize count variable for storing // length of sub-array int count = 1; // Initialize 'prev' variable which // indicates the previous element // while traversing for index 'i' boolean prev = arr[0]; for (int i = 1; i < n; ++i) { // If both elements are same // print elements because alternate // element is not found for current // index if ((arr[i] ^ prev) == false) { // print count and decrement it. while (count > 0) { System.out.print(count-- + " "); } } // Increment count for next element ++count; // Re-initialize previous variable prev = arr[i]; } // If elements are still available after // traversing whole array, print it while (count != 0) { System.out.print(count-- + " "); } } // Driver program public static void main(String args[]) { boolean arr[] = { true, false, true, false, false, true }; int n = arr.length; alternateSubarray(arr, n); } } /*This code is contributed by 29AjayKumar*/ |
Python3
# Python 3 program to calculate longest # alternating sub-array for each index elements # Function to calculate alternating sub-array # for each index of array elements def alternateSubarray(arr, n): # Initialize count variable for # storing length of sub-array count = 1 # Initialize 'prev' variable which # indicates the previous element # while traversing for index 'i' prev = arr[0] for i in range(1, n): # If both elements are same, print # elements because alternate element # is not found for current index if ((arr[i] ^ prev) == 0): # print count and decrement it. while (count): print(count, end = " ") count -= 1 # Increment count for next element count += 1 # Re-initialize previous variable prev = arr[i] # If elements are still available after # traversing whole array, print it while (count): print(count, end = " ") count -= 1 # Driver Code if __name__ == '__main__': arr = [1, 0, 1, 0, 0, 1] n = len(arr) alternateSubarray(arr, n) # This code is contributed by # Surendra_Gangwar |
C#
// C# program to calculate longest alternating // sub-array for each index elements using System; public class Test{ // Function to calculate alternating sub-array // for each index of array elements static void alternateSubarray(bool []arr, int n) { // Initialize count variable for storing // length of sub-array int count = 1; // Initialize 'prev' variable which // indicates the previous element // while traversing for index 'i' bool prev = arr[0]; for (int i = 1; i < n; ++i) { // If both elements are same // print elements because alternate // element is not found for current // index if ((arr[i] ^ prev) == false) { // print count and decrement it. while (count > 0) { Console.Write(count-- + " "); } } // Increment count for next element ++count; // Re-initialize previous variable prev = arr[i]; } // If elements are still available after // traversing whole array, print it while (count != 0) { Console.Write(count-- + " "); } } // Driver program public static void Main() { bool []arr = {true, false, true, false, false, true}; int n = arr.Length; alternateSubarray(arr, n); } } /*This code is contributed by 29AjayKumar*/ |
PHP
<?php // PHP program to calculate longest alternating // sub-array for each index elements // Function to calculate alternating sub-array // for each index of array elements function alternateSubarray($arr, $n) { // Initialize count variable for storing // length of sub-array $count = 1; // Initialize 'prev' variable which // indicates the previous element // while traversing for index 'i' $prev = $arr[0]; for ($i = 1; $i < $n; ++$i) { // If both elements are same // print elements because alternate // element is not found for current // index if (($arr[$i] ^ $prev) == 0) { // print count and decrement it. while ($count) echo $count-- , " "; } // Increment count for next element ++$count; // Re-initialize previous variable $prev = $arr[$i]; } // If elements are still available after // traversing whole array, print it while ($count) echo $count-- , " "; } // Driver Code $arr = array(1, 0, 1, 0, 0, 1); $n = sizeof($arr) ; alternateSubarray($arr, $n); // This code is contributed by jit_t. ?> |
Javascript
<script> // Javascript program to calculate longest // alternating sub-array for each index elements // Function to calculate alternating sub-array // for each index of array elements function alternateSubarray(arr, n) { // Initialize count variable for storing // length of sub-array let count = 1; // Initialize 'prev' variable which // indicates the previous element // while traversing for index 'i' let prev = arr[0]; for(let i = 1; i < n; ++i) { // If both elements are same // print elements because alternate // element is not found for current // index if ((arr[i] ^ prev) == false) { // Print count and decrement it. while (count > 0) { document.write(count-- + " "); } } // Increment count for next element ++count; // Re-initialize previous variable prev = arr[i]; } // If elements are still available after // traversing whole array, print it while (count != 0) { document.write(count-- + " "); } } // Driver code let arr = [ true, false, true, false, false, true ]; let n = arr.length; alternateSubarray(arr, n); // This code is contributed by suresh07 </script> |
Output
4 3 2 1 2 1
Time complexity: O(n)
Auxiliary space: O(1)
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