Minimum number of jumps required to sort numbers placed on a number line
Given two arrays W[] and L[] consisting of N positive integers, where W[i] is initially located at position i on an infinite number line. In each forward jump, W[i] can jump to the position (j + L[i]) from its current position j to any vacant position. The task is to find the minimum number of forward jumps required to sort the array.
Examples:
Input: W[] = {2, 1, 4, 3}, L[] = {4, 1, 2, 4}
Output: 5
Explanation:
Initially, 2 is at position 0, 1 is at position 1, 4 is at position 2, 3 is at position 3 on the number line as: 2 1 4 3
Push number 2 to jump from its current position 0 to position 4, arrangement on the number line: _ 1 4 3 2
Push number 3 to jump from its current position 3 to position 7, arrangement on the number line: _ 1 4 _ 2 _ _ 3
Push number 4 to jump from its current position 2 to position 8, arrangement on the number line: _ 1 _ _ 2 _ _ 3 4Therefore, the total number of jumps required is 1 + 1 + 3 = 5.
Input: W[] = {3, 1, 2}, L[] = {1, 4, 5}
Output: 3
Approach: The given problem can be solved by using the Greedy Approach by minimizing the number of jumps required for the smallest element in the array which is not correctly positioned in every operation and update the number of jumps. Follow the steps below to solve the problem:
- Initialize a variable, say ans as 0 to store the minimum number of jumps required.
- Create a copy of the array W[] and store the elements in sorted order in array arr[].
- Initialize a HashMap pos that stores the current position of the element W[i].
- Traverse the array W[] and update the position of W[i] in pos.
- Traverse the array arr[] in the range [1, N – 1] and perform the following steps:
- Store the position of arr[i] and the position of arr[i – 1] in the array W[] in the variables, say curr and prev respectively.
- If the value of curr is greater than prev, then continue to the next iteration.
- Otherwise, iterate until curr ? prev or curr is already occupied and increment the value of curr by the jump and increment the value of ans by 1.
- Update the position of arr[i] in the HashMap pos to curr.
- After completing the above steps, print the value of ans as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find the minimum number // of jumps required to sort the array void minJumps( int w[], int l[], int n) { // Base Case if (n == 1) { cout << 0; return ; } // Store the required result int ans = 0; // Stores the current position of // elements and their respective // maximum jump unordered_map< int , int > pos, jump; // Used to check if a position is // already taken by another element unordered_map< int , bool > filled; // Stores the sorted array a[] int a[n]; // Traverse the array w[] & update // positions jumps array a[] for ( int i = 0; i < n; i++) { pos[w[i]] = i; filled[i] = true ; jump[w[i]] = l[i]; a[i] = w[i]; } // Sort the array a[] sort(a, a + n); // Traverse the array a[] over // the range [1, N-1] for ( int curr = 1; curr < n; curr++) { // Store the index of current // element and its just smaller // element in array w[] int currElementPos = pos[a[curr]]; int prevElementPos = pos[a[curr - 1]]; if (currElementPos > prevElementPos) continue ; // Iterate until current element // position is at most its just // smaller element position while (currElementPos <= prevElementPos || filled[currElementPos]) { currElementPos += jump[a[curr]]; ans++; } // Update the position of the // current element pos[a[curr]] = currElementPos; filled[currElementPos] = true ; } // Print the result cout << ans; } // Driver Code int main() { int W[] = { 2, 1, 4, 3 }; int L[] = { 4, 1, 2, 4 }; int N = sizeof (W) / sizeof (W[0]); minJumps(W, L, N); return 0; } |
Java
// Java program for the above approach import java.util.*; class GFG { // Function to find the minimum number // of jumps required to sort the array static void minJumps( int [] w, int [] l, int n) { // Base Case if (n == 1 ) { System.out.print( 0 ); return ; } // Store the required result int ans = 0 ; // Stores the current position of // elements and their respective // maximum jump HashMap<Integer, Integer> pos = new HashMap<Integer, Integer>(); HashMap<Integer, Integer> jump = new HashMap<Integer, Integer>(); // Used to check if a position is // already taken by another element HashMap<Integer, Boolean> filled = new HashMap<Integer, Boolean>(); // Stores the sorted array a[] int [] a = new int [n]; // Traverse the array w[] & update // positions jumps array a[] for ( int i = 0 ; i < n; i++) { if (pos.containsKey(w[i])) pos.put(w[i], i); else pos.put(w[i], i); if (filled.containsKey(w[i])) filled.put(i, true ); else filled.put(i, true ); if (jump.containsKey(w[i])) jump.put(w[i], l[i]); else jump.put(w[i], l[i]); a[i] = w[i]; } // Sort the array a[] Arrays.sort(a); // Traverse the array a[] over // the range [1, N-1] for ( int curr = 1 ; curr < n; curr++) { // Store the index of current // element and its just smaller // element in array w[] int currElementPos = pos.get(a[curr]); int prevElementPos = pos.get(a[curr - 1 ]); if (currElementPos > prevElementPos) continue ; // Iterate until current element // position is at most its just // smaller element position while (currElementPos <= prevElementPos || filled.containsKey(currElementPos) && filled.containsKey( currElementPos)) { currElementPos += jump.get(a[curr]); ans++; } // Update the position of the // current element if (pos.containsKey(a[curr])) pos.put(a[curr], currElementPos); else pos.put(a[curr], currElementPos); if (filled.containsKey(currElementPos)) filled.put(currElementPos, true ); else filled.put(currElementPos, true ); } // Print the result System.out.print(ans); } // Driver Code public static void main(String[] args) { int [] W = { 2 , 1 , 4 , 3 }; int [] L = { 4 , 1 , 2 , 4 }; int N = W.length; minJumps(W, L, N); } } // This code is contributed by ukasp. |
Python3
# Python3 program for the above approach # Function to find the minimum number # of jumps required to sort the array def minJumps(w, l, n): # Base Case if (n = = 1 ): print ( 0 ) return # Store the required result ans = 0 # Stores the current position of # elements and their respective # maximum jump pos = {} jump = {} # Used to check if a position is # already taken by another element filled = {} # Stores the sorted array a[] a = [ 0 for i in range (n)] # Traverse the array w[] & update # positions jumps array a[] for i in range (n): pos[w[i]] = i filled[i] = True jump[w[i]] = l[i] a[i] = w[i] # Sort the array a[] a.sort() # Traverse the array a[] over # the range [1, N-1] for curr in range ( 1 , n, 1 ): # Store the index of current # element and its just smaller # element in array w[] currElementPos = pos[a[curr]] prevElementPos = pos[a[curr - 1 ]] if (currElementPos > prevElementPos): continue # Iterate until current element # position is at most its just # smaller element position while (currElementPos < = prevElementPos or (currElementPos in filled and filled[currElementPos])): currElementPos + = jump[a[curr]] ans + = 1 # Update the position of the # current element pos[a[curr]] = currElementPos filled[currElementPos] = True # Print the result print (ans) # Driver Code if __name__ = = '__main__' : W = [ 2 , 1 , 4 , 3 ] L = [ 4 , 1 , 2 , 4 ] N = len (W) minJumps(W, L, N) # This code is contributed by SURENDRA_GANGWAR |
C#
// C# program for the above approach using System; using System.Collections.Generic; class GFG{ // Function to find the minimum number // of jumps required to sort the array static void minJumps( int []w, int []l, int n) { // Base Case if (n == 1) { Console.Write(0); return ; } // Store the required result int ans = 0; // Stores the current position of // elements and their respective // maximum jump Dictionary< int , int > pos = new Dictionary< int , int >(); Dictionary< int , int > jump = new Dictionary< int , int >(); // Used to check if a position is // already taken by another element Dictionary< int , bool > filled = new Dictionary< int , bool >(); // Stores the sorted array a[] int []a = new int [n]; // Traverse the array w[] & update // positions jumps array a[] for ( int i = 0; i < n; i++) { if (pos.ContainsKey(w[i])) pos[w[i]] = i; else pos.Add(w[i], i); if (filled.ContainsKey(w[i])) filled[i] = true ; else filled.Add(i, true ); if (jump.ContainsKey(w[i])) jump[w[i]] = l[i]; else jump.Add(w[i], l[i]); a[i] = w[i]; } // Sort the array a[] Array.Sort(a); // Traverse the array a[] over // the range [1, N-1] for ( int curr = 1; curr < n; curr++) { // Store the index of current // element and its just smaller // element in array w[] int currElementPos = pos[a[curr]]; int prevElementPos = pos[a[curr - 1]]; if (currElementPos > prevElementPos) continue ; // Iterate until current element // position is at most its just // smaller element position while (currElementPos <= prevElementPos || filled.ContainsKey(currElementPos) && filled[currElementPos]) { currElementPos += jump[a[curr]]; ans++; } // Update the position of the // current element if (pos.ContainsKey(a[curr])) pos[a[curr]] = currElementPos; else pos.Add(a[curr], currElementPos); if (filled.ContainsKey(currElementPos)) filled[currElementPos] = true ; else filled.Add(currElementPos, true ); } // Print the result Console.Write(ans); } // Driver Code public static void Main() { int []W = { 2, 1, 4, 3 }; int []L = { 4, 1, 2, 4 }; int N = W.Length; minJumps(W, L, N); } } // This code is contributed by ipg2016107 |
Javascript
<script> // Javascript program for the above approach // Function to find the minimum number // of jumps required to sort the array function minJumps(w, l, n) { // Base Case if (n == 1) { document.write(0); return ; } // Store the required result var ans = 0; var i; // Stores the current position of // elements and their respective // maximum jump var pos = new Map(); var jump = new Map(); // Used to check if a position is // already taken by another element var filled = new Map(); // Stores the sorted array a[] var a = new Array(n); // Traverse the array w[] & update // positions jumps array a[] for (i = 0; i < n; i++) { pos.set(w[i], i); filled.set(i, true ); jump.set(w[i], l[i]); a[i] = w[i]; } // Sort the array a[] a = a.sort( function (p, q) { return p - q; }); // Traverse the array a[] over // the range [1, N-1] for (curr = 1; curr < n; curr++) { // Store the index of current // element and its just smaller // element in array w[] var currElementPos = pos.get(a[curr]); var prevElementPos = pos.get(a[curr - 1]); if (currElementPos > prevElementPos) continue ; // Iterate until current element // position is at most its just // smaller element position while (currElementPos <= prevElementPos || filled[currElementPos]) { currElementPos += jump.get(a[curr]); ans += 1; } // Update the position of the // current element pos.set(a[curr], currElementPos); filled.set(currElementPos, true ); } // Print the result document.write(ans); } // Driver Code var W = [ 2, 1, 4, 3 ]; var L = [ 4, 1, 2, 4 ]; var N = W.length; minJumps(W, L, N); // This code is contributed by ipg2016107 </script> |
5
Time Complexity: O(N*log N)
Auxiliary Space: O(N)
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