Given an array representing n positions along a straight line. Find k (where k <= n) elements from the array such that the minimum distance between any two (consecutive points among the k points) is maximized.
Examples :
Input: arr[] = {1, 2, 8, 4, 9}, k = 3
Output: 3
Explanation: Largest minimum distance = 3
3 elements arranged at positions 1, 4 and 8,
Resulting in a minimum distance of 3Input: arr[] = {1, 2, 7, 5, 11, 12}, k = 3
Output: 5
Explanation: 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 the 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 a binary search for maximum result for given k. We start with the middle of the maximum possible result. If the middle is a feasible solution, we search on the right half of mid. Else we search is left half. To check feasibility, we place k elements under given mid-distance.
Implementation:
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
//here we are using right value as highest distance difference,
//so we remove some extra checks
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 if
// 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;
} |
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 = 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 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. |
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 = 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):
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 = 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 |
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 = 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 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. |
Javascript
<script> // Javascript 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
let pos = arr[0];
// Initialize count of elements placed.
let elements = 1;
// Try placing k elements with minimum
// distance mid.
for (let 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.
function largestMinDist(arr, n, k)
{
// Sort the positions
arr.sort(function(a, b){return a - b});
// Initialize result.
let res = -1;
// Consider the maximum possible
// distance
let 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) {
let mid = parseInt((left + right) / 2, 10);
// 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;
}
let arr = [ 1, 2, 8, 4, 9 ];
let n = arr.length;
let k = 3;
document.write(largestMinDist(arr, n, k));
</script> |
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 = 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)
{
$mid = ($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
$res = max($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 ?> |
Output
3
Time Complexity: O(n log m) , where n is length of array and m is the maximum element of the array.
Auxiliary Space: O(1)