Find the minimum of maximum length of a jump required to reach the last island in exactly k jumps
Given an array arr[] of integers, where the ith integer represents the position where an island is present, and an integer k (1 ≤ k < N). A person is standing on the 0th island and has to reach the last island, by jumping from one island to another in exactly k jumps, the task is to find the minimum of the maximum length of a jump a person will make in his journey. Note that the position of all the islands are given in ascending order.
Examples:
Input: arr[] = {2, 15, 36, 43}, k = 1
Output: 41
There is only way to reach the end
2 -> 43
Input: arr[] = {2, 15, 36, 43}, k = 2
Output: 28
There are two ways to reach the last island
2 -> 15 -> 43
Here maximum distance between any two consecutive islands is between 43 and 15 that is 28.
2 -> 36 -> 43
Here maximum distance between any two consecutive islands is between 36 and 2 that is 34.
Thus minimum of 28 and 34 is 28.
Approach: The idea is to use binary search, and for a distance mid, compute whether it is possible to reach the end of the array in exactly k jumps where the maximum distance between any two islands chosen for jumping is less than or equal to the distance mid, then check if some distance less than mid exists for which it is possible to reach the end in exactly k jumps.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function that returns true if it is possible // to reach end of the array in exactly k jumps bool isPossible(int arr[], int n, int dist, int k) { // Variable to store the number of // steps required to reach the end int req = 0; int curr = 0; int prev = 0; for (int i = 0; i < n; i++) { while (curr != n && arr[curr] - arr[prev] <= dist) curr++; req++; if (curr == n) break; prev = curr - 1; } if (curr != n) return false; // If it is possible to reach the // end in exactly k jumps if (req <= k) return true; return false; } // Returns the minimum maximum distance required // to reach the end of the array in exactly k jumps int minDistance(int arr[], int n, int k) { int l = 0; int h = arr[n - 1]; // Stores the answer int ans = 0; // Binary search to calculate the result while (l <= h) { int m = (l + h) / 2; if (isPossible(arr, n, m, k)) { ans = m; h = m - 1; } else l = m + 1; } return ans; } // Driver code int main() { int arr[] = { 2, 15, 36, 43 }; int n = sizeof(arr) / sizeof(int); int k = 2; cout << minDistance(arr, n, k); return 0; } |
chevron_right
filter_none
Java
// Java implementation of the approach class GFG { // Function that returns true if it is possible // to reach end of the array in exactly k jumps static boolean isPossible(int arr[], int n, int dist, int k) { // Variable to store the number of // steps required to reach the end int req = 0; int curr = 0; int prev = 0; for (int i = 0; i < n; i++) { while (curr != n && arr[curr] - arr[prev] <= dist) { curr++; } req++; if (curr == n) { break; } prev = curr - 1; } if (curr != n) { return false; } // If it is possible to reach the // end in exactly k jumps if (req <= k) { return true; } return false; } // Returns the minimum maximum distance required // to reach the end of the array in exactly k jumps static int minDistance(int arr[], int n, int k) { int l = 0; int h = arr[n - 1]; // Stores the answer int ans = 0; // Binary search to calculate the result while (l <= h) { int m = (l + h) / 2; if (isPossible(arr, n, m, k)) { ans = m; h = m - 1; } else { l = m + 1; } } return ans; } // Driver code public static void main(String[] args) { int arr[] = {2, 15, 36, 43}; int n = arr.length; int k = 2; System.out.println(minDistance(arr, n, k)); } } /* This code contributed by PrinciRaj1992 */ |
chevron_right
filter_none
Python3
# Python3 implementation of the approach # Function that returns true if it is possible # to reach end of the array in exactly k jumps def isPossible(arr, n, dist, k) : # Variable to store the number of # steps required to reach the end req = 0 curr = 0 prev = 0 for i in range(0, n): while (curr != n and (arr[curr] - arr[prev]) <= dist): curr = curr + 1 req = req + 1 if (curr == n): break prev = curr - 1 if (curr != n): return False # If it is possible to reach the # end in exactly k jumps if (req <= k): return True return False # Returns the minimum maximum distance required # to reach the end of the array in exactly k jumps def minDistance(arr, n, k): l = 0 h = arr[n - 1] # Stores the answer ans = 0 # Binary search to calculate the result while (l <= h): m = (l + h) // 2; if (isPossible(arr, n, m, k)): ans = m h = m - 1 else: l = m + 1 return ans # Driver code arr = [ 2, 15, 36, 43 ] n = len(arr) k = 2 print(minDistance(arr, n, k)) # This code is contributed by ihritik |
chevron_right
filter_none
C#
// C# program to implement // the above approach using System; class GFG { // Function that returns true if it is possible // to reach end of the array in exactly k jumps static bool isPossible(int []arr, int n, int dist, int k) { // Variable to store the number of // steps required to reach the end int req = 0; int curr = 0; int prev = 0; for (int i = 0; i < n; i++) { while (curr != n && arr[curr] - arr[prev] <= dist) { curr++; } req++; if (curr == n) { break; } prev = curr - 1; } if (curr != n) { return false; } // If it is possible to reach the // end in exactly k jumps if (req <= k) { return true; } return false; } // Returns the minimum maximum distance required // to reach the end of the array in exactly k jumps static int minDistance(int []arr, int n, int k) { int l = 0; int h = arr[n - 1]; // Stores the answer int ans = 0; // Binary search to calculate the result while (l <= h) { int m = (l + h) / 2; if (isPossible(arr, n, m, k)) { ans = m; h = m - 1; } else { l = m + 1; } } return ans; } // Driver code public static void Main(String[] args) { int []arr = {2, 15, 36, 43}; int n = arr.Length; int k = 2; Console.WriteLine(minDistance(arr, n, k)); } } /* This code contributed by PrinciRaj1992 */ |
chevron_right
filter_none
PHP
<?php // Php implementation of the approach // Function that returns true if it is possible // to reach end of the array in exactly k jumps function isPossible($arr, $n, $dist, $k) { // Variable to store the number of // steps required to reach the end $req = 0; $curr = 0; $prev = 0; for ($i = 0; $i < $n; $i++) { while ($curr != $n && $arr[$curr] - $arr[$prev] <= $dist) $curr++; $req++; if ($curr == $n) break; $prev = $curr - 1; } if ($curr != $n) return false; // If it is possible to reach the // end in exactly k jumps if ($req <= $k) return true; return false; } // Returns the minimum maximum distance required // to reach the end of the array in exactly k jumps function minDistance($arr, $n, $k) { $l = 0; $h = $arr[$n - 1]; // Stores the answer $ans = 0; // Binary search to calculate the result while ($l <= $h) { $m = floor(($l + $h) / 2); if (isPossible($arr, $n, $m, $k)) { $ans = $m; $h = $m - 1; } else $l = $m + 1; } return $ans; } // Driver code $arr = array( 2, 15, 36, 43 ); $n = count($arr); $k = 2; echo minDistance($arr, $n, $k); // This code is contributed by Ryuga ?> |
chevron_right
filter_none
Output:
28
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
