# Min and max length subarray having adjacent element difference atmost K

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}

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ program to find the minimum ` `// and maximum length subarray such ` `// that difference between adjacent ` `// elements is atmost K ` ` `  `#include ` ` `  `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 ` `    ``// associateed 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); ` ` `  `    ``// function call to find maximum ` `    ``// and minimum possible sub array ` `    ``findMaxMinSubArray(arr, K, n); ` ` `  `    ``return` `0; ` `} `

 `// 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 ` `    ``// associateed 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 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  ` `    ``# associateed 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# 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 ` `    ``// associateed 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 `

Output:
```2, 3
```

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