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# Min difference between maximum and minimum element in all Y size subarrays

• Difficulty Level : Medium
• Last Updated : 10 Jun, 2021

Given an array arr[] of size N and integer Y, the task is to find a minimum of all the differences between the maximum and minimum elements in all the sub-arrays of size Y.

Examples:

Input: arr[] = { 3, 2, 4, 5, 6, 1, 9 }  Y = 3
Output: 2
Explanation:
All subarrays of length = 3 are:
{3, 2, 4} where maximum element = 4 and  minimum element = 2  difference = 2
{2, 4, 5} where maximum element = 5 and  minimum element = 2  difference = 3
{4, 5, 6} where maximum element = 6 and  minimum element = 4  difference = 2
{5, 6, 1} where maximum element = 6 and  minimum element = 1  difference = 5
{6, 1, 9} where maximum element = 9 and  minimum element = 6  difference = 3
Out of these, the minimum is 2.

Input: arr[] = { 1, 2, 3, 3, 2, 2  } Y = 4
Output: 1
Explanation:
All subarrays of length = 4 are:
{1, 2, 3, 3} maximum element = 3 and  minimum element = 1  difference = 2
{2, 3, 3, 2} maximum element = 3 and  minimum element = 2  difference = 1
{3, 3, 2, 2} maximum element = 3 and  minimum element = 2  difference = 1
Out of these, the minimum is 1.

Naive Approach: The naive idea is to traverse for every index i in the range [0, N – Y] use another loop to traverse from ith index to (i + Y – 1)th index and then calculate the minimum and maximum elements in a subarray of size Y and hence calculate the difference of maximum and minimum element for that ith iteration. Finally, by keeping a check on differences, evaluate the minimum difference.

Time Complexity: O(N*Y)
Auxiliary Space: O(1)

Efficient Approach: The idea is to use the concept of the approach discussed in the Next Greater Element article. Below are the steps:

1. Build two arrays maxarr[] and minarr[], where maxarr[] will store the index of the element which is next greater to the element at ith index and minarr[] will store the index of the next element which is less than the element at the ith index.
2. Initialize a stack with 0 to store the indices in both the above cases.
3. For each index, i in the range [0, N – Y], using a nested loop and sliding window approach form two arrays submax and submin. These arrays will store maximum and minimum elements in the subarray in ith iteration.
4. Finally, calculate the minimum difference.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to get the maximum of all``// the subarrays of size Y``vector<``int``> get_submaxarr(``int``* arr,``                        ``int` `n, ``int` `y)``{``    ``int` `j = 0;``    ``stack<``int``> stk;` `    ``// ith index of maxarr array``    ``// will be the index upto which``    ``// Arr[i] is maximum``    ``vector<``int``> maxarr(n);``    ``stk.push(0);` `    ``for` `(``int` `i = 1; i < n; i++) {` `        ``// Stack is used to find the``        ``// next larger element and``        ``// keeps track of index of``        ``// current iteration``        ``while` `(stk.empty() == ``false``            ``and arr[i] > arr[stk.top()]) {` `            ``maxarr[stk.top()] = i - 1;``            ``stk.pop();``        ``}``        ``stk.push(i);``    ``}` `    ``// Loop for remaining indexes``    ``while` `(!stk.empty()) {` `        ``maxarr[stk.top()] = n - 1;``        ``stk.pop();``    ``}``    ``vector<``int``> submax;` `    ``for` `(``int` `i = 0; i <= n - y; i++) {` `        ``// j < i used to keep track``        ``// whether jth element is``        ``// inside or outside the window``        ``while` `(maxarr[j] < i + y - 1``            ``or j < i) {``            ``j++;``        ``}` `        ``submax.push_back(arr[j]);``    ``}` `    ``// Return submax``    ``return` `submax;``}` `// Function to get the minimum for``// all subarrays of size Y``vector<``int``> get_subminarr(``int``* arr,``                        ``int` `n, ``int` `y)``{``    ``int` `j = 0;` `    ``stack<``int``> stk;` `    ``// ith index of minarr array``    ``// will be the index upto which``    ``// Arr[i] is minimum``    ``vector<``int``> minarr(n);``    ``stk.push(0);` `    ``for` `(``int` `i = 1; i < n; i++) {` `        ``// Stack is used to find the``        ``// next smaller element and``        ``// keeping track of index of``        ``// current iteration``        ``while` `(stk.empty() == ``false``            ``and arr[i] < arr[stk.top()]) {` `            ``minarr[stk.top()] = i;``            ``stk.pop();``        ``}``        ``stk.push(i);``    ``}` `    ``// Loop for remaining indexes``    ``while` `(!stk.empty()) {` `        ``minarr[stk.top()] = n;``        ``stk.pop();``    ``}` `    ``vector<``int``> submin;` `    ``for` `(``int` `i = 0; i <= n - y; i++) {` `        ``// j < i used to keep track``        ``// whether jth element is inside``        ``// or outside the window` `        ``while` `(minarr[j] <= i + y - 1``            ``or j < i) {``            ``j++;``        ``}` `        ``submin.push_back(arr[j]);``    ``}` `    ``// Return submin``    ``return` `submin;``}` `// Function to get minimum difference``void` `getMinDifference(``int` `Arr[], ``int` `N,``                    ``int` `Y)``{``    ``// Create submin and submax arrays``    ``vector<``int``> submin``        ``= get_subminarr(Arr, N, Y);` `    ``vector<``int``> submax``        ``= get_submaxarr(Arr, N, Y);` `    ``// Store initial difference``    ``int` `minn = submax - submin;``    ``int` `b = submax.size();` `    ``for` `(``int` `i = 1; i < b; i++) {` `        ``// Calculate temporary difference``        ``int` `dif = submax[i] - submin[i];``        ``minn = min(minn, dif);``    ``}` `    ``// Final minimum difference``    ``cout << minn << ``"\n"``;``}` `// Driver Code``int` `main()``{``    ``// Given array arr[]``    ``int` `arr[] = { 1, 2, 3, 3, 2, 2 };``    ``int` `N = ``sizeof` `arr / ``sizeof` `arr;` `    ``// Given subarray size``    ``int` `Y = 4;` `    ``// Function Call``    ``getMinDifference(arr, N, Y);``    ``return` `0;``}`

## Python3

 `# Python3 program for the above approach` `# Function to get the maximum of all``# the subarrays of size Y``def` `get_submaxarr(arr, n, y):``    ` `    ``j ``=` `0``    ``stk ``=` `[]``    ` `    ``# ith index of maxarr array``    ``# will be the index upto which``    ``# Arr[i] is maximum``    ``maxarr ``=` `[``0``] ``*` `n``    ``stk.append(``0``)` `    ``for` `i ``in` `range``(``1``, n):` `        ``# Stack is used to find the``        ``# next larger element and``        ``# keeps track of index of``        ``# current iteration``        ``while` `(``len``(stk) > ``0` `and``                 ``arr[i] > arr[stk[``-``1``]]):``            ``maxarr[stk[``-``1``]] ``=` `i ``-` `1``            ``stk.pop()``            ` `        ``stk.append(i)` `    ``# Loop for remaining indexes``    ``while` `(stk):``        ``maxarr[stk[``-``1``]] ``=` `n ``-` `1``        ``stk.pop()``        ` `    ``submax ``=` `[]``    ` `    ``for` `i ``in` `range``(n ``-` `y ``+` `1``):` `        ``# j < i used to keep track``        ``# whether jth element is``        ``# inside or outside the window``        ``while` `(maxarr[j] < i ``+` `y ``-` `1` `or``                       ``j < i):``            ``j ``+``=` `1``            ` `        ``submax.append(arr[j])` `    ``# Return submax``    ``return` `submax` `# Function to get the minimum for``# all subarrays of size Y``def` `get_subminarr(arr, n, y):``    ` `    ``j ``=` `0``    ``stk ``=` `[]``    ` `    ``# ith index of minarr array``    ``# will be the index upto which``    ``# Arr[i] is minimum``    ``minarr ``=` `[``0``] ``*` `n``    ``stk.append(``0``)``    ` `    ``for` `i ``in` `range``(``1` `, n):``        ` `        ``# Stack is used to find the``        ``# next smaller element and``        ``# keeping track of index of``        ``# current iteration``        ``while` `(stk ``and` `arr[i] < arr[stk[``-``1``]]):``            ``minarr[stk[``-``1``]] ``=` `i``            ``stk.pop()``            ` `        ``stk.append(i)``        ` `    ``# Loop for remaining indexes``    ``while` `(stk):``        ``minarr[stk[``-``1``]] ``=` `n``        ``stk.pop()``        ` `    ``submin ``=` `[]``    ` `    ``for` `i ``in` `range``(n ``-` `y ``+` `1``):``        ` `        ``# j < i used to keep track``        ``# whether jth element is inside``        ``# or outside the window``        ``while` `(minarr[j] <``=` `i ``+` `y ``-` `1` `or``                      ``j < i):``            ``j ``+``=` `1``            ` `        ``submin.append(arr[j])``        ` `    ``# Return submin``    ``return` `submin` `# Function to get minimum difference``def` `getMinDifference(Arr, N, Y):``    ` `    ``# Create submin and submax arrays``    ``submin ``=` `get_subminarr(Arr, N, Y)``    ``submax ``=` `get_submaxarr(Arr, N, Y)``    ` `    ``# Store initial difference``    ``minn ``=` `submax[``0``] ``-` `submin[``0``]``    ``b ``=` `len``(submax)``    ` `    ``for` `i ``in` `range``(``1``, b):``        ` `        ``# Calculate temporary difference``        ``diff ``=` `submax[i] ``-` `submin[i]``        ``minn ``=` `min``(minn, diff)``    ` `    ``# Final minimum difference``    ``print``(minn)``    ` `# Driver code` `# Given array arr[]``arr ``=` `[ ``1``, ``2``, ``3``, ``3``, ``2``, ``2` `]``N ``=` `len``(arr)` `# Given subarray size``Y ``=` `4` `# Function call``getMinDifference(arr, N, Y)``    ` `# This code is contributed by Stuti Pathak`

## Javascript

 ``
Output:
`1`

Time Complexity: O(N)
Auxiliary Space: O(N)

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