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++ 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 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 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# 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 |
<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)