Given an array representing n positions along a straight line. Find k (where k <= n) elements from the array such that the maximum distance between any two (consecutive points among the k points) is maximized.**Examples :**

Input : arr[] = {1, 2, 8, 4, 9} k = 3 Output : 3 Largest minimum distance = 3 3 elements arranged at positions 1, 4 and 8, Resulting in a minimum distance of 3 Input : arr[] = {1, 2, 7, 5, 11, 12} k = 3 Output : 5 Largest minimum distance = 5 3 elements arranged at positions 1, 7 and 12, resulting in a minimum distance of 5 (between 7 and 12)

A **Naive Solution** is to consider all subsets of size 3 and find minimum distance for every subset. Finally return the largest of all minimum distances.

An **Efficient Solution** is based on Binary Search. We first sort the array. Now we know maximum possible value result is arr[n-1] – arr[0] (for k = 2). We do binary search for maximum result for given k. We start with middle of maximum possible result. If middle is a feasible solution, we search on right half of mid. Else we search is left half. To check feasibility, we place k elements under given mid distance.

## C++

`// C++ program to find largest minimum distance` `// among k points.` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Returns true if it is possible to arrange` `// k elements of arr[0..n-1] with minimum distance` `// given as mid.` `bool` `isFeasible(` `int` `mid, ` `int` `arr[], ` `int` `n, ` `int` `k)` `{` ` ` `// Place first element at arr[0] position` ` ` `int` `pos = arr[0];` ` ` `// Initialize count of elements placed.` ` ` `int` `elements = 1;` ` ` `// Try placing k elements with minimum` ` ` `// distance mid.` ` ` `for` `(` `int` `i=1; i<n; i++)` ` ` `{` ` ` `if` `(arr[i] - pos >= mid)` ` ` `{` ` ` `// Place next element if its` ` ` `// distance from the previously` ` ` `// placed element is greater` ` ` `// than current mid` ` ` `pos = arr[i];` ` ` `elements++;` ` ` `// Return if all elements are placed` ` ` `// successfully` ` ` `if` `(elements == k)` ` ` `return` `true` `;` ` ` `}` ` ` `}` ` ` `return` `0;` `}` `// Returns largest minimum distance for k elements` `// in arr[0..n-1]. If elements can't be placed,` `// returns -1.` `int` `largestMinDist(` `int` `arr[], ` `int` `n, ` `int` `k)` `{` ` ` `// Sort the positions` ` ` `sort(arr,arr+n);` ` ` `// Initialize result.` ` ` `int` `res = -1;` ` ` `// Consider the maximum possible distance` ` ` `int` `left = 1, right = arr[n-1];` ` ` ` ` `// left is initialized with 1 and not with arr[0] ` ` ` `// because, minimum distance between each element ` ` ` `// can be one and not arr[0]. consider this example: ` ` ` `// arr[] = {9,12} and you have to place 2 element` ` ` `// then left = arr[0] will force the function to` ` ` `// look the answer between range arr[0] to arr[n-1], ` ` ` `// i.e 9 to 12, but the answer is 3 so It is required ` ` ` `// that you initialize the left with 1` ` ` `// Do binary search for largest minimum distance` ` ` `while` `(left < right)` ` ` `{` ` ` `int` `mid = (left + right)/2;` ` ` `// If it is possible to place k elements` ` ` `// with minimum distance mid, search for` ` ` `// higher distance.` ` ` `if` `(isFeasible(mid, arr, n, k))` ` ` `{` ` ` `// Change value of variable max to mid iff` ` ` `// all elements can be successfully placed` ` ` `res = max(res, mid);` ` ` `left = mid + 1;` ` ` `}` ` ` `// If not possible to place k elements, search` ` ` `// for lower distance` ` ` `else` ` ` `right = mid;` ` ` `}` ` ` `return` `res;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = {1, 2, 8, 4, 9};` ` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]);` ` ` `int` `k = 3;` ` ` `cout << largestMinDist(arr, n, k);` ` ` `return` `0;` `}` |

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## Java

`// Java program to find largest ` `// minimum distance among k points.` `import` `java.util.Arrays;` `class` `GFG ` `{` `// Returns true if it is possible to` `// arrange k elements of arr[0..n-1]` `// with minimum distance given as mid.` `static` `boolean` `isFeasible(` `int` `mid, ` `int` `arr[], ` ` ` `int` `n, ` `int` `k)` `{` ` ` `// Place first element at arr[0] position` ` ` `int` `pos = arr[` `0` `];` ` ` `// Initialize count of elements placed.` ` ` `int` `elements = ` `1` `;` ` ` `// Try placing k elements with minimum` ` ` `// distance mid.` ` ` `for` `(` `int` `i=` `1` `; i<n; i++)` ` ` `{` ` ` `if` `(arr[i] - pos >= mid)` ` ` `{` ` ` `// Place next element if its` ` ` `// distance from the previously` ` ` `// placed element is greater` ` ` `// than current mid` ` ` `pos = arr[i];` ` ` `elements++;` ` ` `// Return if all elements are ` ` ` `// placed successfully` ` ` `if` `(elements == k)` ` ` `return` `true` `;` ` ` `}` ` ` `}` ` ` `return` `false` `;` `}` `// Returns largest minimum distance for` `// k elements in arr[0..n-1]. If elements ` `// can't be placed, returns -1.` `static` `int` `largestMinDist(` `int` `arr[], ` `int` `n, ` ` ` `int` `k)` `{` ` ` `// Sort the positions` ` ` `Arrays.sort(arr);` ` ` `// Initialize result.` ` ` `int` `res = -` `1` `;` ` ` `// Consider the maximum possible distance` ` ` `int` `left = arr[` `0` `], right = arr[n-` `1` `];` ` ` ` ` `// left is initialized with 1 and not with arr[0] ` ` ` `// because, minimum distance between each element ` ` ` `// can be one and not arr[0]. consider this example: ` ` ` `// arr[] = {9,12} and you have to place 2 element` ` ` `// then left = arr[0] will force the function to` ` ` `// look the answer between range arr[0] to arr[n-1], ` ` ` `// i.e 9 to 12, but the answer is 3 so It is required ` ` ` `// that you initialize the left with 1` ` ` `// Do binary search for largest` ` ` `// minimum distance` ` ` `while` `(left < right)` ` ` `{` ` ` `int` `mid = (left + right)/` `2` `;` ` ` `// If it is possible to place k ` ` ` `// elements with minimum distance mid, ` ` ` `// search for higher distance.` ` ` `if` `(isFeasible(mid, arr, n, k))` ` ` `{` ` ` `// Change value of variable max to` ` ` `// mid if all elements can be` ` ` `// successfully placed` ` ` `res = Math.max(res, mid);` ` ` `left = mid + ` `1` `;` ` ` `}` ` ` `// If not possible to place k elements, ` ` ` `// search for lower distance` ` ` `else` ` ` `right = mid;` ` ` `}` ` ` `return` `res;` `}` `// driver code` `public` `static` `void` `main (String[] args)` `{` ` ` `int` `arr[] = {` `1` `, ` `2` `, ` `8` `, ` `4` `, ` `9` `};` ` ` `int` `n = arr.length;` ` ` `int` `k = ` `3` `;` ` ` `System.out.print(largestMinDist(arr, n, k));` `}` `}` `// This code is contributed by Anant Agarwal.` |

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## Python3

`# Python 3 program to find largest minimum ` `# distance among k points.` `# Returns true if it is possible to arrange` `# k elements of arr[0..n-1] with minimum ` `# distance given as mid.` `def` `isFeasible(mid, arr, n, k):` ` ` ` ` `# Place first element at arr[0] position` ` ` `pos ` `=` `arr[` `0` `]` ` ` `# Initialize count of elements placed.` ` ` `elements ` `=` `1` ` ` `# Try placing k elements with minimum` ` ` `# distance mid.` ` ` `for` `i ` `in` `range` `(` `1` `, n, ` `1` `):` ` ` `if` `(arr[i] ` `-` `pos >` `=` `mid):` ` ` ` ` `# Place next element if its distance ` ` ` `# from the previously placed element` ` ` `# is greater than current mid` ` ` `pos ` `=` `arr[i]` ` ` `elements ` `+` `=` `1` ` ` `# Return if all elements are placed` ` ` `# successfully` ` ` `if` `(elements ` `=` `=` `k):` ` ` `return` `True` ` ` `return` `0` `# Returns largest minimum distance for k elements` `# in arr[0..n-1]. If elements can't be placed,` `# returns -1.` `def` `largestMinDist(arr, n, k):` ` ` ` ` `# Sort the positions` ` ` `arr.sort(reverse ` `=` `False` `)` ` ` `# Initialize result.` ` ` `res ` `=` `-` `1` ` ` `# Consider the maximum possible distance` ` ` `left ` `=` `arr[` `0` `]` ` ` `right ` `=` `arr[n ` `-` `1` `]` ` ` ` ` `# left is initialized with 1 and not with arr[0] ` ` ` `# because, minimum distance between each element ` ` ` `# can be one and not arr[0]. consider this example: ` ` ` `# arr[] = {9,12} and you have to place 2 element` ` ` `# then left = arr[0] will force the function to` ` ` `# look the answer between range arr[0] to arr[n-1], ` ` ` `# i.e 9 to 12, but the answer is 3 so It is required ` ` ` `# that you initialize the left with 1` ` ` `# Do binary search for largest ` ` ` `# minimum distance` ` ` `while` `(left < right):` ` ` `mid ` `=` `(left ` `+` `right) ` `/` `2` ` ` `# If it is possible to place k elements` ` ` `# with minimum distance mid, search for` ` ` `# higher distance.` ` ` `if` `(isFeasible(mid, arr, n, k)):` ` ` ` ` `# Change value of variable max to mid iff` ` ` `# all elements can be successfully placed` ` ` `res ` `=` `max` `(res, mid)` ` ` `left ` `=` `mid ` `+` `1` ` ` `# If not possible to place k elements, ` ` ` `# search for lower distance` ` ` `else` `:` ` ` `right ` `=` `mid` ` ` `return` `res` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `arr ` `=` `[` `1` `, ` `2` `, ` `8` `, ` `4` `, ` `9` `]` ` ` `n ` `=` `len` `(arr)` ` ` `k ` `=` `3` ` ` `print` `(largestMinDist(arr, n, k))` `# This code is contributed by` `# Sanjit_prasad` |

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## C#

`// C# program to find largest ` `// minimum distance among k points.` `using` `System;` `public` `class` `GFG {` ` ` ` ` `// Returns true if it is possible to` ` ` `// arrange k elements of arr[0..n-1]` ` ` `// with minimum distance given as mid.` ` ` `static` `bool` `isFeasible(` `int` `mid, ` `int` `[]arr, ` ` ` `int` `n, ` `int` `k)` ` ` `{` ` ` ` ` `// Place first element at arr[0]` ` ` `// position` ` ` `int` `pos = arr[0];` ` ` ` ` `// Initialize count of elements placed.` ` ` `int` `elements = 1;` ` ` ` ` `// Try placing k elements with minimum` ` ` `// distance mid.` ` ` `for` `(` `int` `i = 1; i < n; i++)` ` ` `{` ` ` `if` `(arr[i] - pos >= mid)` ` ` `{` ` ` ` ` `// Place next element if its` ` ` `// distance from the previously` ` ` `// placed element is greater` ` ` `// than current mid` ` ` `pos = arr[i];` ` ` `elements++;` ` ` ` ` `// Return if all elements are ` ` ` `// placed successfully` ` ` `if` `(elements == k)` ` ` `return` `true` `;` ` ` `}` ` ` `}` ` ` ` ` `return` `false` `;` ` ` `}` ` ` ` ` `// Returns largest minimum distance for` ` ` `// k elements in arr[0..n-1]. If elements ` ` ` `// can't be placed, returns -1.` ` ` `static` `int` `largestMinDist(` `int` `[]arr, ` `int` `n, ` ` ` `int` `k)` ` ` `{` ` ` ` ` `// Sort the positions` ` ` `Array.Sort(arr);` ` ` ` ` `// Initialize result.` ` ` `int` `res = -1;` ` ` ` ` `// Consider the maximum possible` ` ` `// distance` ` ` `int` `left = arr[0], right = arr[n-1];` ` ` ` ` `// left is initialized with 1 and not with arr[0] ` ` ` `// because, minimum distance between each element ` ` ` `// can be one and not arr[0]. consider this example: ` ` ` `// arr[] = {9,12} and you have to place 2 element` ` ` `// then left = arr[0] will force the function to` ` ` `// look the answer between range arr[0] to arr[n-1], ` ` ` `// i.e 9 to 12, but the answer is 3 so It is required ` ` ` `// that you initialize the left with 1` ` ` ` ` `// Do binary search for largest` ` ` `// minimum distance` ` ` `while` `(left < right)` ` ` `{` ` ` `int` `mid = (left + right) / 2;` ` ` ` ` `// If it is possible to place k ` ` ` `// elements with minimum distance` ` ` `// mid, search for higher distance.` ` ` `if` `(isFeasible(mid, arr, n, k))` ` ` `{` ` ` `// Change value of variable` ` ` `// max to mid if all elements` ` ` `// can be successfully placed` ` ` `res = Math.Max(res, mid);` ` ` `left = mid + 1;` ` ` `}` ` ` ` ` `// If not possible to place k` ` ` `// elements, search for lower` ` ` `// distance` ` ` `else` ` ` `right = mid;` ` ` `}` ` ` ` ` `return` `res;` ` ` `}` ` ` ` ` `// driver code` ` ` `public` `static` `void` `Main ()` ` ` `{` ` ` `int` `[]arr = {1, 2, 8, 4, 9};` ` ` `int` `n = arr.Length;` ` ` `int` `k = 3;` ` ` ` ` `Console.WriteLine(` ` ` `largestMinDist(arr, n, k));` ` ` `}` `}` `// This code is contributed by Sam007.` |

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## PHP

`<?php` `// PHP program to find largest ` `// minimum distance among k points.` `// Returns true if it is possible ` `// to arrange k elements of ` `// arr[0..n-1] with minimum ` `// distance given as mid.` `function` `isFeasible(` `$mid` `, ` `$arr` `, ` ` ` `$n` `, ` `$k` `)` `{` ` ` `// Place first element ` ` ` `// at arr[0] position` ` ` `$pos` `= ` `$arr` `[0];` ` ` `// Initialize count of` ` ` `// elements placed.` ` ` `$elements` `= 1;` ` ` `// Try placing k elements ` ` ` `// with minimum distance mid.` ` ` `for` `(` `$i` `= 1; ` `$i` `< ` `$n` `; ` `$i` `++)` ` ` `{` ` ` `if` `(` `$arr` `[` `$i` `] - ` `$pos` `>= ` `$mid` `)` ` ` `{` ` ` `// Place next element if ` ` ` `// its distance from the ` ` ` `// previously placed ` ` ` `// element is greater ` ` ` `// than current mid` ` ` `$pos` `= ` `$arr` `[` `$i` `];` ` ` `$elements` `++;` ` ` `// Return if all elements ` ` ` `// are placed successfully` ` ` `if` `(` `$elements` `== ` `$k` `)` ` ` `return` `true;` ` ` `}` ` ` `}` ` ` `return` `0;` `}` `// Returns largest minimum ` `// distance for k elements ` `// in arr[0..n-1]. If elements ` `// can't be placed, returns -1.` `function` `largestMinDist(` `$arr` `, ` `$n` `, ` `$k` `)` `{` ` ` `// Sort the positions` ` ` `sort(` `$arr` `);` ` ` `// Initialize result.` ` ` `$res` `= -1;` ` ` `// Consider the maximum` ` ` `// possible distance` ` ` `$left` `= ` `$arr` `[0];` ` ` `$right` `= ` `$arr` `[` `$n` `- 1];` ` ` ` ` `// left is initialized with 1 and not with arr[0] ` ` ` `// because, minimum distance between each element ` ` ` `// can be one and not arr[0]. consider this example: ` ` ` `// arr[] = {9,12} and you have to place 2 element` ` ` `// then left = arr[0] will force the function to` ` ` `// look the answer between range arr[0] to arr[n-1], ` ` ` `// i.e 9 to 12, but the answer is 3 so It is required ` ` ` `// that you initialize the left with 1` ` ` `// Do binary search for` ` ` `// largest minimum distance` ` ` `while` `(` `$left` `< ` `$right` `)` ` ` `{` ` ` `$mid` `= (` `$left` `+ ` `$right` `) / 2;` ` ` `// If it is possible to place ` ` ` `// k elements with minimum ` ` ` `// distance mid, search for ` ` ` `// higher distance.` ` ` `if` `(isFeasible(` `$mid` `, ` `$arr` `, ` ` ` `$n` `, ` `$k` `))` ` ` `{` ` ` `// Change value of variable ` ` ` `// max to mid iff all elements` ` ` `// can be successfully placed` ` ` `$res` `= max(` `$res` `, ` `$mid` `);` ` ` `$left` `= ` `$mid` `+ 1;` ` ` `}` ` ` `// If not possible to place ` ` ` `// k elements, search for` ` ` `// lower distance` ` ` `else` ` ` `$right` `= ` `$mid` `;` ` ` `}` ` ` `return` `$res` `;` `}` `// Driver Code` `$arr` `= ` `array` `(1, 2, 8, 4, 9);` `$n` `= sizeof(` `$arr` `);` `$k` `= 3;` `echo` `largestMinDist(` `$arr` `, ` `$n` `, ` `$k` `);` `// This code is contributed by aj_36` `?>` |

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**Output :**

3

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