Given an array arr[] and an integer K, the task is to find the maximum and minimum length subarray such that the difference between adjacent elements is at most K.
Examples:
Input: arr[] = {2, 4, 6}, K = 2
Output: 3, 3
Explanation:
Minimum and Maximum length subarray such that difference
between adjacent elements is at most is 3, which is {2, 4, 2}
Input: arr[] = {2, 3, 6, 7, 8}, K = 2
Output: 2, 3
Explanation:
Minimum length Subarray(2) => {2, 3}
Maximum length Subarray(3) => {6, 7, 8}
Approach: The idea is traverse the array, to start from each element and move to its right and left direction until the difference between the adjacent elements is less than K. Finally, update the maximum and minimum length subarray with the current length as defined below –
if (current_length maximum_length) maximum_length = current_length
Below is the implementation of the above approach:
C++
// C++ program to find the minimum // and maximum length subarray such // that difference between adjacent // elements is atmost K #include <iostream> using namespace std; // Function to find the maximum and // minimum length subarray void findMaxMinSubArray( int arr[], int K, int n) { // Initialise minimum and maximum // size of subarray in worst case int min = n; int max = 0; // Left will scan the size of // possible subarray in left // of selected element int left; // Right will scan the size of // possible subarray in right // of selected element int right; // Temp will store size of group // associated with every element int tmp; // Loop to find size of subarray // for every element of array for ( int i = 0; i < n; i++) { tmp = 1; left = i; // Left will move in left direction // and compare difference between // itself and element left to it while (left - 1 >= 0 && abs (arr[left] - arr[left - 1]) <= K) { left--; tmp++; } // right will move in right direction // and compare difference between // itself and element right to it right = i; while (right + 1 <= n - 1 && abs (arr[right] - arr[right + 1]) <= K) { right++; tmp++; } // if subarray of much lesser or much // greater is found than yet // known then update if (min > tmp) min = tmp; if (max < tmp) max = tmp; } // Print minimum and maximum // possible size of subarray cout << min << ", " << max << endl; } // Driver Code int main() { int arr[] = { 1, 2, 5, 6, 7 }; int K = 2; int n = sizeof (arr) / sizeof (arr[0]); // function call to find maximum // and minimum possible sub array findMaxMinSubArray(arr, K, n); return 0; } |
Java
// Java program to find the minimum // and maximum length subarray such // that difference between adjacent // elements is atmost K import java.io.*; import java.util.*; class GFG { // Function to find the maximum and // minimum length subarray static void findMaxMinSubArray( int arr[], int K, int n) { // Initialise minimum and maximum // size of subarray in worst case int min = n; int max = 0 ; // Left will scan the size of // possible subarray in left // of selected element int left; // Right will scan the size of // possible subarray in right // of selected element int right; // Temp will store size of group // associated with every element int tmp; // Loop to find size of subarray // for every element of array for ( int i = 0 ; i < n; i++) { tmp = 1 ; left = i; // Left will move in left direction // and compare difference between // itself and element left to it while (left - 1 >= 0 && Math.abs(arr[left] - arr[left - 1 ]) <= K) { left--; tmp++; } // Right will move in right direction // and compare difference between // itself and element right to it right = i; while (right + 1 <= n - 1 && Math.abs(arr[right] - arr[right + 1 ]) <= K) { right++; tmp++; } // If subarray of much lesser or much // greater is found than yet // known then update if (min > tmp) min = tmp; if (max < tmp) max = tmp; } // Print minimum and maximum // possible size of subarray System.out.print(min); System.out.print( ", " ); System.out.print(max); } // Driver code public static void main(String[] args) { int arr[] = { 1 , 2 , 5 , 6 , 7 }; int K = 2 ; int n = arr.length; // Function call to find maximum // and minimum possible sub array findMaxMinSubArray(arr, K, n); } } // This code is contributed by coder001 |
Python3
# Python3 program to find the minimum # and maximum length subarray such # that difference between adjacent # elements is atmost K # Function to find the maximum and # minimum length subarray def findMaxMinSubArray(arr, K, n): # Initialise minimum and maximum # size of subarray in worst case min = n max = 0 # Left will scan the size of # possible subarray in left # of selected element left = 0 # Right will scan the size of # possible subarray in right # of selected element right = n # Temp will store size of group # associated with every element tmp = 0 # Loop to find size of subarray # for every element of array for i in range ( 0 , n): tmp = 1 left = i # Left will move in left direction # and compare difference between # itself and element left to it while (left - 1 > = 0 and abs (arr[left] - arr[left - 1 ]) < = K): left = left - 1 tmp = tmp + 1 # Right will move in right direction # and compare difference between # itself and element right to it right = i while (right + 1 < = n - 1 and abs (arr[right] - arr[right + 1 ]) < = K): right = right + 1 tmp = tmp + 1 # If subarray of much lesser or much # greater is found than yet # known then update if ( min > tmp): min = tmp if ( max < tmp): max = tmp # Print minimum and maximum # possible size of subarray print ( min , end = ', ' ) print ( max , end = '\n' ) # Driver Code arr = [ 1 , 2 , 5 , 6 , 7 ] K = 2 n = len (arr) # Function call to find maximum # and minimum possible sub array findMaxMinSubArray(arr, K, n) # This code is contributed by Pratik |
C#
// C# program to find the minimum // and maximum length subarray such // that difference between adjacent // elements is atmost K using System; class GFG{ // Function to find the maximum and // minimum length subarray static void findMaxMinSubArray( int [] arr, int K, int n) { // Initialise minimum and maximum // size of subarray in worst case int min = n; int max = 0; // Left will scan the size of // possible subarray in left // of selected element int left; // Right will scan the size of // possible subarray in right // of selected element int right; // Temp will store size of group // associated with every element int tmp; // Loop to find size of subarray // for every element of array for ( int i = 0; i < n; i++) { tmp = 1; left = i; // Left will move in left direction // and compare difference between // itself and element left to it while (left - 1 >= 0 && Math.Abs(arr[left] - arr[left - 1]) <= K) { left--; tmp++; } // Right will move in right direction // and compare difference between // itself and element right to it right = i; while (right + 1 <= n - 1 && Math.Abs(arr[right] - arr[right + 1]) <= K) { right++; tmp++; } // If subarray of much lesser or much // greater is found than yet // known then update if (min > tmp) min = tmp; if (max < tmp) max = tmp; } // Print minimum and maximum // possible size of subarray Console.Write(min); Console.Write( ", " ); Console.Write(max); } // Driver code public static void Main() { int [] arr = { 1, 2, 5, 6, 7 }; int K = 2; int n = arr.Length; // Function call to find maximum // and minimum possible sub array findMaxMinSubArray(arr, K, n); } } // This code is contributed by AbhiThakur |
Javascript
<script> // Java Script program to find the minimum // and maximum length subarray such // that difference between adjacent // elements is atmost K // Function to find the maximum and // minimum length subarray function findMaxMinSubArray(arr, k, n) { // Initialise minimum and maximum // size of subarray in worst case let min = n; let max = 0; // Left will scan the size of // possible subarray in left // of selected element let left; // Right will scan the size of // possible subarray in right // of selected element let right; // Temp will store size of group // associated with every element let tmp; // Loop to find size of subarray // for every element of array for (let i = 0; i < n; i++) { tmp = 1; left = i; // Left will move in left direction // and compare difference between // itself and element left to it while (left - 1 >= 0 && Math.abs(arr[left] - arr[left - 1]) <= K) { left--; tmp++; } // Right will move in right direction // and compare difference between // itself and element right to it right = i; while (right + 1 <= n - 1 && Math.abs(arr[right] - arr[right + 1]) <= K) { right++; tmp++; } // If subarray of much lesser or much // greater is found than yet // known then update if (min > tmp) min = tmp; if (max < tmp) max = tmp; } // Print minimum and maximum // possible size of subarray document.write(min); document.write( ", " ); document.write(max); } // Driver code let arr = [1, 2, 5, 6, 7 ]; let K = 2; let n = arr.length; // Function call to find maximum // and minimum possible sub array findMaxMinSubArray(arr, K, n); // This code is contributed by sravan </script> |
2, 3
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