Program for minimum number of jumps to reach the end
Here given code implementation process.
/*
C program for
Program for minimum number of jumps to reach the end
*/
#include <stdio.h>
#include <limits.h>
// This is finding the number of minimum steps
// or jumps required to reach end of array elements
// When of array not contains negative values
int minimumJumpsToEnd(int arr[], int n)
{
if (n <= 1)
{
// When less than 2 elements
return 0;
}
if (arr[0] == 0)
{
// When first element is 0 then not possible to
// Visit next element
// Therefor
return -1;
}
// Use to collect number of jump process
int jump[n];
// Initial value
jump[0] = 0;
// Execute loop through by size of n
for (int i = 1; i < n; ++i)
{
// Set initial steps jump
jump[i] = INT_MAX;
// Inner loop which is finding minimum jump between 0 to i
for (int j = 0; j < i; ++j)
{
if (i <= j + arr[j] && jump[j] != INT_MAX)
{
if (jump[i] > (jump[j] + 1))
{
// Update new minimum jump
jump[i] = jump[j] + 1;
}
// Stop inner loop execution
j = i;
}
}
}
// When jump[ n-1] is INT_MAX
// That means we not reach the end of array elements
return jump[n - 1];
}
void displayArray(int arr[], int n)
{
for (int i = 0; i < n; ++i)
{
printf(" %d", arr[i]);
}
printf("\n");
}
int main(int argc, char const *argv[])
{
int arr1[] = {
2 , 1 , 3 , 2 , 3 , 1 , 6 , 7 , 8 , 4 , 2 , 9
};
int arr2[] = {
2 , 3 , 0 , 1 , 2 , 3 , 0 , 1 , 2 , 6 , 4
};
// Test A
int n = sizeof(arr1) / sizeof(arr1[0]);
displayArray(arr1, n);
// Given [2, 1, 3, 2, 3, 1 , 6, 7, 8, 4, 2, 9 ]
// Jump start with 2 -> 3 -> 1 -> 6 -> 9
// Jump between ① ② ③ ④
// Ans : 4
printf(" Minimum Jump to end : %d\n", minimumJumpsToEnd(arr1, n));
// Test B
n = sizeof(arr2) / sizeof(arr2[0]);
displayArray(arr2, n);
// Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
// Jump start with 2 -> 3 -> 2 -> 3 -> 2 -> 4
// Jump between ① ② ③ ④ ⑤
printf(" Minimum Jump to end : %d\n", minimumJumpsToEnd(arr2, n));
return 0;
}
Output
2 1 3 2 3 1 6 7 8 4 2 9
Minimum Jump to end : 4
2 3 0 1 2 3 0 1 2 6 4
Minimum Jump to end : 5
// Java Program
// Program for minimum number of jumps to reach the end
public class Jumping
{
// This is finding the number of minimum steps
// or jumps required to reach end of array elements
// When of array not contains negative values
public int minimumJumpsToEnd(int[] arr, int n)
{
if (n <= 1)
{
// When less than 2 elements
return 0;
}
if (arr[0] == 0)
{
// When first element is 0 then not possible to
// Visit next element
// Therefor
return -1;
}
// Use to collect number of jump process
int[] jump = new int[n];
// Initial value
jump[0] = 0;
// Execute loop through by size of n
for (int i = 1; i < n; ++i)
{
// Set initial steps jump
jump[i] = Integer.MAX_VALUE;
// Inner loop which is finding minimum jump between 0 to i
for (int j = 0; j < i; ++j)
{
if (i <= j + arr[j] && jump[j] != Integer.MAX_VALUE)
{
if (jump[i] > (jump[j] + 1))
{
// Update new minimum jump
jump[i] = jump[j] + 1;
}
// Stop inner loop execution
j = i;
}
}
}
// When jump[ n-1] is MAX Integer
// That means we not reach the end of array elements
return jump[n - 1];
}
public void displayArray(int[] arr, int n)
{
for (int i = 0; i < n; ++i)
{
System.out.print(" " + arr[i] );
}
System.out.print("\n");
}
public static void main(String args[])
{
Jumping task = new Jumping();
int[] arr1 = {
2 , 1 , 3 , 2 , 3 , 1 , 6 , 7 , 8 , 4 , 2 , 9
};
int[] arr2 = {
2 , 3 , 0 , 1 , 2 , 3 , 0 , 1 , 2 , 6 , 4
};
// Test A
int n = arr1.length;
task.displayArray(arr1, n);
// Given [2, 1, 3, 2, 3, 1 , 6, 7, 8, 4, 2, 9 ]
// Jump start with 2 → 3 → 1 → 6 → 9
// Jump between ① ② ③ ④
// Ans : 4
System.out.println(" Minimum Jump to end : " +
task.minimumJumpsToEnd(arr1, n) );
// Test B
n = arr2.length;
task.displayArray(arr2, n);
// Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
// Jump start with 2 → 3 → 2 → 3 → 2 → 4
// Jump between ① ② ③ ④ ⑤
System.out.println(" Minimum Jump to end : " +
task.minimumJumpsToEnd(arr2, n) );
}
}
Output
2 1 3 2 3 1 6 7 8 4 2 9
Minimum Jump to end : 4
2 3 0 1 2 3 0 1 2 6 4
Minimum Jump to end : 5
// Include header file
#include <iostream>
#include <limits.h>
using namespace std;
// C++ Program
// Program for minimum number of jumps to reach the end
class Jumping
{
public:
// This is finding the number of minimum steps
// or jumps required to reach end of array elements
// When of array not contains negative values
int minimumJumpsToEnd(int arr[], int n)
{
if (n <= 1)
{
// When less than 2 elements
return 0;
}
if (arr[0] == 0)
{
// When first element is 0 then not possible to
// Visit next element
// Therefor
return -1;
}
// Use to collect number of jump process
int jump[n];
// Initial value
jump[0] = 0;
// Execute loop through by size of n
for (int i = 1; i < n; ++i)
{
// Set initial steps jump
jump[i] = INT_MAX;
// Inner loop which is finding minimum jump between 0 to i
for (int j = 0; j < i; ++j)
{
if (i <= j + arr[j] && jump[j] != INT_MAX)
{
if (jump[i] > (jump[j] + 1))
{
// Update new minimum jump
jump[i] = jump[j] + 1;
}
// Stop inner loop execution
j = i;
}
}
}
// When jump[ n-1] is MAX Integer
// That means we not reach the end of array elements
return jump[n - 1];
}
void displayArray(int arr[], int n)
{
for (int i = 0; i < n; ++i)
{
cout << " " << arr[i];
}
cout << "\n";
}
};
int main()
{
Jumping *task = new Jumping();
int arr1[] = {
2 , 1 , 3 , 2 , 3 , 1 , 6 , 7 , 8 , 4 , 2 , 9
};
int arr2[] = {
2 , 3 , 0 , 1 , 2 , 3 , 0 , 1 , 2 , 6 , 4
};
// Test A
int n = sizeof(arr1) / sizeof(arr1[0]);
task->displayArray(arr1, n);
// Given [2, 1, 3, 2, 3, 1 , 6, 7, 8, 4, 2, 9 ]
// Jump start with 2 → 3 → 1 → 6 → 9
// Jump between ① ② ③ ④
// Ans : 4
cout << " Minimum Jump to end : " << task->minimumJumpsToEnd(arr1, n) << endl;
// Test B
n = sizeof(arr2) / sizeof(arr2[0]);
task->displayArray(arr2, n);
// Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
// Jump start with 2 → 3 → 2 → 3 → 2 → 4
// Jump between ① ② ③ ④ ⑤
cout << " Minimum Jump to end : " << task->minimumJumpsToEnd(arr2, n) << endl;
return 0;
}
Output
2 1 3 2 3 1 6 7 8 4 2 9
Minimum Jump to end : 4
2 3 0 1 2 3 0 1 2 6 4
Minimum Jump to end : 5
// Include namespace system
using System;
// Csharp Program
// Program for minimum number of jumps to reach the end
public class Jumping
{
// This is finding the number of minimum steps
// or jumps required to reach end of array elements
// When of array not contains negative values
public int minimumJumpsToEnd(int[] arr, int n)
{
if (n <= 1)
{
// When less than 2 elements
return 0;
}
if (arr[0] == 0)
{
// When first element is 0 then not possible to
// Visit next element
// Therefor
return -1;
}
// Use to collect number of jump process
int[] jump = new int[n];
// Initial value
jump[0] = 0;
// Execute loop through by size of n
for (int i = 1; i < n; ++i)
{
// Set initial steps jump
jump[i] = int.MaxValue;
// Inner loop which is finding minimum jump between 0 to i
for (int j = 0; j < i; ++j)
{
if (i <= j + arr[j] && jump[j] != int.MaxValue)
{
if (jump[i] > (jump[j] + 1))
{
// Update new minimum jump
jump[i] = jump[j] + 1;
}
// Stop inner loop execution
j = i;
}
}
}
// When jump[ n-1] is MAX Integer
// That means we not reach the end of array elements
return jump[n - 1];
}
public void displayArray(int[] arr, int n)
{
for (int i = 0; i < n; ++i)
{
Console.Write(" " + arr[i]);
}
Console.Write("\n");
}
public static void Main(String[] args)
{
Jumping task = new Jumping();
int[] arr1 = {
2 , 1 , 3 , 2 , 3 , 1 , 6 , 7 , 8 , 4 , 2 , 9
};
int[] arr2 = {
2 , 3 , 0 , 1 , 2 , 3 , 0 , 1 , 2 , 6 , 4
};
// Test A
int n = arr1.Length;
task.displayArray(arr1, n);
// Given [2, 1, 3, 2, 3, 1 , 6, 7, 8, 4, 2, 9 ]
// Jump start with 2 → 3 → 1 → 6 → 9
// Jump between ① ② ③ ④
// Ans : 4
Console.WriteLine(" Minimum Jump to end : " +
task.minimumJumpsToEnd(arr1, n));
// Test B
n = arr2.Length;
task.displayArray(arr2, n);
// Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
// Jump start with 2 → 3 → 2 → 3 → 2 → 4
// Jump between ① ② ③ ④ ⑤
Console.WriteLine(" Minimum Jump to end : " +
task.minimumJumpsToEnd(arr2, n));
}
}
Output
2 1 3 2 3 1 6 7 8 4 2 9
Minimum Jump to end : 4
2 3 0 1 2 3 0 1 2 6 4
Minimum Jump to end : 5
package main
import "math"
import "fmt"
// Go Program
// Program for minimum number of jumps to reach the end
// This is finding the number of minimum steps
// or jumps required to reach end of array elements
// When of array not contains negative values
func minimumJumpsToEnd(arr[] int, n int) int {
if n <= 1 {
// When less than 2 elements
return 0
}
if arr[0] == 0 {
// When first element is 0 then not possible to
// Visit next element
// Therefor
return -1
}
// Use to collect number of jump process
var jump = make([]int,n)
// Execute loop through by size of n
for i := 1 ; i < n ; i++ {
// Set initial steps jump
jump[i] = math.MaxInt64
// Inner loop which is finding minimum jump between 0 to i
for j := 0 ; j < i ; j++ {
if i <= j + arr[j] && jump[j] != math.MaxInt64 {
if jump[i] > (jump[j] + 1) {
// Update new minimum jump
jump[i] = jump[j] + 1
}
// Stop inner loop execution
j = i
}
}
}
// When jump[ n-1] is MAX Integer
// That means we not reach the end of array elements
return jump[n - 1]
}
func displayArray(arr[] int, n int) {
for i := 0 ; i < n ; i++ {
fmt.Print(" ", arr[i])
}
fmt.Print("\n")
}
func main() {
var arr1 = [] int {2, 1, 3, 2, 3, 1 , 6, 7, 8, 4, 2, 9 }
var arr2 = [] int { 2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4 }
// Test A
var n int = len(arr1)
displayArray(arr1, n)
// Given [2, 1, 3, 2, 3, 1 , 6, 7, 8, 4, 2, 9 ]
// Jump start with 2 → 3 → 1 → 6 → 9
// Jump between ① ② ③ ④
// Ans : 4
fmt.Println(" Minimum Jump to end : ", minimumJumpsToEnd(arr1, n))
// Test B
n = len(arr2)
displayArray(arr2, n)
// Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
// Jump start with 2 → 3 → 2 → 3 → 2 → 4
// Jump between ① ② ③ ④ ⑤
fmt.Println(" Minimum Jump to end : ", minimumJumpsToEnd(arr2, n))
}
Output
2 1 3 2 3 1 6 7 8 4 2 9
Minimum Jump to end : 4
2 3 0 1 2 3 0 1 2 6 4
Minimum Jump to end : 5
<?php
// Php Program
// Program for minimum number of jumps to reach the end
class Jumping
{
// This is finding the number of minimum steps
// or jumps required to reach end of array elements
// When of array not contains negative values
public function minimumJumpsToEnd($arr, $n)
{
if ($n <= 1)
{
// When less than 2 elements
return 0;
}
if ($arr[0] == 0)
{
// When first element is 0 then not possible to
// Visit next element
// Therefor
return -1;
}
// Use to collect number of jump process
$jump = array_fill(0, $n, 0);
// Execute loop through by size of n
for ($i = 1; $i < $n; ++$i)
{
// Set initial steps jump
$jump[$i] = PHP_INT_MAX;
// Inner loop which is finding minimum jump between 0 to i
for ($j = 0; $j < $i; ++$j)
{
if ($i <= $j + $arr[$j] && $jump[$j] != PHP_INT_MAX)
{
if ($jump[$i] > ($jump[$j] + 1))
{
// Update new minimum jump
$jump[$i] = $jump[$j] + 1;
}
// Stop inner loop execution
$j = $i;
}
}
}
// When jump[ n-1] is MAX Integer
// That means we not reach the end of array elements
return $jump[$n - 1];
}
public function displayArray($arr, $n)
{
for ($i = 0; $i < $n; ++$i)
{
echo(" ".$arr[$i]);
}
echo("\n");
}
}
function main()
{
$task = new Jumping();
$arr1 = array(2, 1, 3, 2, 3, 1, 6, 7, 8, 4, 2, 9);
$arr2 = array(2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4);
// Test A
$n = count($arr1);
$task->displayArray($arr1, $n);
// Given [2, 1, 3, 2, 3, 1 , 6, 7, 8, 4, 2, 9 ]
// Jump start with 2 → 3 → 1 → 6 → 9
// Jump between ① ② ③ ④
// Ans : 4
echo(" Minimum Jump to end : ".
$task->minimumJumpsToEnd($arr1, $n).
"\n");
// Test B
$n = count($arr2);
$task->displayArray($arr2, $n);
// Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
// Jump start with 2 → 3 → 2 → 3 → 2 → 4
// Jump between ① ② ③ ④ ⑤
echo(" Minimum Jump to end : ".
$task->minimumJumpsToEnd($arr2, $n).
"\n");
}
main();
Output
2 1 3 2 3 1 6 7 8 4 2 9
Minimum Jump to end : 4
2 3 0 1 2 3 0 1 2 6 4
Minimum Jump to end : 5
// Node JS Program
// Program for minimum number of jumps to reach the end
class Jumping
{
// This is finding the number of minimum steps
// or jumps required to reach end of array elements
// When of array not contains negative values
minimumJumpsToEnd(arr, n)
{
if (n <= 1)
{
// When less than 2 elements
return 0;
}
if (arr[0] == 0)
{
// When first element is 0 then not possible to
// Visit next element
// Therefor
return -1;
}
// Use to collect number of jump process
var jump = Array(n).fill(0);
// Execute loop through by size of n
for (var i = 1; i < n; ++i)
{
// Set initial steps jump
jump[i] = Number.MAX_VALUE;
// Inner loop which is finding minimum jump between 0 to i
for (var j = 0; j < i; ++j)
{
if (i <= j + arr[j] && jump[j] != Number.MAX_VALUE)
{
if (jump[i] > (jump[j] + 1))
{
// Update new minimum jump
jump[i] = jump[j] + 1;
}
// Stop inner loop execution
j = i;
}
}
}
// When jump[ n-1] is MAX Integer
// That means we not reach the end of array elements
return jump[n - 1];
}
displayArray(arr, n)
{
for (var i = 0; i < n; ++i)
{
process.stdout.write(" " + arr[i]);
}
process.stdout.write("\n");
}
}
function main()
{
var task = new Jumping();
var arr1 = [2, 1, 3, 2, 3, 1, 6, 7, 8, 4, 2, 9];
var arr2 = [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4];
// Test A
var n = arr1.length;
task.displayArray(arr1, n);
// Given [2, 1, 3, 2, 3, 1 , 6, 7, 8, 4, 2, 9 ]
// Jump start with 2 → 3 → 1 → 6 → 9
// Jump between ① ② ③ ④
// Ans : 4
console.log(" Minimum Jump to end : " +
task.minimumJumpsToEnd(arr1, n));
// Test B
n = arr2.length;
task.displayArray(arr2, n);
// Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
// Jump start with 2 → 3 → 2 → 3 → 2 → 4
// Jump between ① ② ③ ④ ⑤
console.log(" Minimum Jump to end : " +
task.minimumJumpsToEnd(arr2, n));
}
main();
Output
2 1 3 2 3 1 6 7 8 4 2 9
Minimum Jump to end : 4
2 3 0 1 2 3 0 1 2 6 4
Minimum Jump to end : 5
import sys
# Python 3 Program
# Program for minimum number of jumps to reach the end
class Jumping :
# This is finding the number of minimum steps
# or jumps required to reach end of list elements
# When of list not contains negative values
def minimumJumpsToEnd(self, arr, n) :
if (n <= 1) :
# When less than 2 elements
return 0
if (arr[0] == 0) :
# When first element is 0 then not possible to
# Visit next element
# Therefor
return -1
# Use to collect number of jump process
jump = [0] * (n)
i = 1
# Execute loop through by size of n
while (i < n) :
# Set initial steps jump
jump[i] = sys.maxsize
j = 0
# Inner loop which is finding minimum jump between 0 to i
while (j < i) :
if (i <= j + arr[j] and jump[j] != sys.maxsize) :
if (jump[i] > (jump[j] + 1)) :
# Update new minimum jump
jump[i] = jump[j] + 1
# Stop inner loop execution
j = i
j += 1
i += 1
# When jump[ n-1] is MAX Integer
# That means we not reach the end of list elements
return jump[n - 1]
def displayArray(self, arr, n) :
i = 0
while (i < n) :
print(" ", arr[i], end = "")
i += 1
print(end = "\n")
def main() :
task = Jumping()
arr1 = [2, 1, 3, 2, 3, 1, 6, 7, 8, 4, 2, 9]
arr2 = [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
# Test A
n = len(arr1)
task.displayArray(arr1, n)
# Given [2, 1, 3, 2, 3, 1 , 6, 7, 8, 4, 2, 9 ]
# Jump start with 2 → 3 → 1 → 6 → 9
# Jump between ① ② ③ ④
# Ans : 4
print(" Minimum Jump to end : ",
task.minimumJumpsToEnd(arr1, n))
# Test B
n = len(arr2)
task.displayArray(arr2, n)
# Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
# Jump start with 2 → 3 → 2 → 3 → 2 → 4
# Jump between ① ② ③ ④ ⑤
print(" Minimum Jump to end : ",
task.minimumJumpsToEnd(arr2, n))
if __name__ == "__main__": main()
Output
2 1 3 2 3 1 6 7 8 4 2 9
Minimum Jump to end : 4
2 3 0 1 2 3 0 1 2 6 4
Minimum Jump to end : 5
# Ruby Program
# Program for minimum number of jumps to reach the end
class Jumping
# This is finding the number of minimum steps
# or jumps required to reach end of array elements
# When of array not contains negative values
def minimumJumpsToEnd(arr, n)
if (n <= 1)
# When less than 2 elements
return 0
end
if (arr[0] == 0)
# When first element is 0 then not possible to
# Visit next element
# Therefor
return -1
end
# Use to collect number of jump process
jump = Array.new(n) {0}
i = 1
# Execute loop through by size of n
while (i < n)
# Set initial steps jump
jump[i] = (2 ** (0. size * 8 - 2))
j = 0
# Inner loop which is finding minimum jump between 0 to i
while (j < i)
if (i <= j + arr[j] && jump[j] != (2 ** (0. size * 8 - 2)))
if (jump[i] > (jump[j] + 1))
# Update new minimum jump
jump[i] = jump[j] + 1
end
# Stop inner loop execution
j = i
end
j += 1
end
i += 1
end
# When jump[ n-1] is MAX Integer
# That means we not reach the end of array elements
return jump[n - 1]
end
def displayArray(arr, n)
i = 0
while (i < n)
print(" ", arr[i])
i += 1
end
print("\n")
end
end
def main()
task = Jumping.new()
arr1 = [2, 1, 3, 2, 3, 1, 6, 7, 8, 4, 2, 9]
arr2 = [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
# Test A
n = arr1.length
task.displayArray(arr1, n)
# Given [2, 1, 3, 2, 3, 1 , 6, 7, 8, 4, 2, 9 ]
# Jump start with 2 → 3 → 1 → 6 → 9
# Jump between ① ② ③ ④
# Ans : 4
print(" Minimum Jump to end : ",
task.minimumJumpsToEnd(arr1, n), "\n")
# Test B
n = arr2.length
task.displayArray(arr2, n)
# Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
# Jump start with 2 → 3 → 2 → 3 → 2 → 4
# Jump between ① ② ③ ④ ⑤
print(" Minimum Jump to end : ",
task.minimumJumpsToEnd(arr2, n), "\n")
end
main()
Output
2 1 3 2 3 1 6 7 8 4 2 9
Minimum Jump to end : 4
2 3 0 1 2 3 0 1 2 6 4
Minimum Jump to end : 5
// Scala Program
// Program for minimum number of jumps to reach the end
class Jumping()
{
// This is finding the number of minimum steps
// or jumps required to reach end of array elements
// When of array not contains negative values
def minimumJumpsToEnd(arr: Array[Int], n: Int): Int = {
if (n <= 1)
{
// When less than 2 elements
return 0;
}
if (arr(0) == 0)
{
// When first element is 0 then not possible to
// Visit next element
// Therefor
return -1;
}
// Use to collect number of jump process
var jump: Array[Int] = Array.fill[Int](n)(0);
var i: Int = 1;
// Execute loop through by size of n
while (i < n)
{
// Set initial steps jump
jump(i) = Int.MaxValue;
var j: Int = 0;
// Inner loop which is finding minimum jump between 0 to i
while (j < i)
{
if (i <= j + arr(j) && jump(j) != Int.MaxValue)
{
if (jump(i) > (jump(j) + 1))
{
// Update new minimum jump
jump(i) = jump(j) + 1;
}
// Stop inner loop execution
j = i;
}
j += 1;
}
i += 1;
}
// When jump[ n-1] is MAX Integer
// That means we not reach the end of array elements
return jump(n - 1);
}
def displayArray(arr: Array[Int], n: Int): Unit = {
var i: Int = 0;
while (i < n)
{
print(" " + arr(i));
i += 1;
}
print("\n");
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: Jumping = new Jumping();
var arr1: Array[Int] = Array(2, 1, 3, 2, 3, 1, 6, 7, 8, 4, 2, 9);
var arr2: Array[Int] = Array(2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4);
// Test A
var n: Int = arr1.length;
task.displayArray(arr1, n);
// Given [2, 1, 3, 2, 3, 1 , 6, 7, 8, 4, 2, 9 ]
// Jump start with 2 → 3 → 1 → 6 → 9
// Jump between ① ② ③ ④
// Ans : 4
println(" Minimum Jump to end : " +
task.minimumJumpsToEnd(arr1, n));
// Test B
n = arr2.length;
task.displayArray(arr2, n);
// Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
// Jump start with 2 → 3 → 2 → 3 → 2 → 4
// Jump between ① ② ③ ④ ⑤
println(" Minimum Jump to end : " +
task.minimumJumpsToEnd(arr2, n));
}
}
Output
2 1 3 2 3 1 6 7 8 4 2 9
Minimum Jump to end : 4
2 3 0 1 2 3 0 1 2 6 4
Minimum Jump to end : 5
import Foundation;
// Swift 4 Program
// Program for minimum number of jumps to reach the end
class Jumping
{
// This is finding the number of minimum steps
// or jumps required to reach end of array elements
// When of array not contains negative values
func minimumJumpsToEnd(_ arr: [Int], _ n: Int) -> Int
{
if (n <= 1)
{
// When less than 2 elements
return 0;
}
if (arr[0] == 0)
{
// When first element is 0 then not possible to
// Visit next element
// Therefor
return -1;
}
// Use to collect number of jump process
var jump: [Int] = Array(repeating: 0, count: n);
var i: Int = 1;
// Execute loop through by size of n
while (i < n)
{
// Set initial steps jump
jump[i] = Int.max;
var j: Int = 0;
// Inner loop which is finding minimum jump between 0 to i
while (j < i)
{
if (i <= j + arr[j] && jump[j] != Int.max)
{
if (jump[i] > (jump[j] + 1))
{
// Update new minimum jump
jump[i] = jump[j] + 1;
}
// Stop inner loop execution
j = i;
}
j += 1;
}
i += 1;
}
// When jump[ n-1] is MAX Integer
// That means we not reach the end of array elements
return jump[n - 1];
}
func displayArray(_ arr: [Int], _ n: Int)
{
var i: Int = 0;
while (i < n)
{
print(" ", arr[i], terminator: "");
i += 1;
}
print(terminator: "\n");
}
}
func main()
{
let task: Jumping = Jumping();
let arr1: [Int] = [2, 1, 3, 2, 3, 1, 6, 7, 8, 4, 2, 9];
let arr2: [Int] = [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4];
// Test A
var n: Int = arr1.count;
task.displayArray(arr1, n);
// Given [2, 1, 3, 2, 3, 1 , 6, 7, 8, 4, 2, 9 ]
// Jump start with 2 → 3 → 1 → 6 → 9
// Jump between ① ② ③ ④
// Ans : 4
print(" Minimum Jump to end : ", task.minimumJumpsToEnd(arr1, n));
// Test B
n = arr2.count;
task.displayArray(arr2, n);
// Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
// Jump start with 2 → 3 → 2 → 3 → 2 → 4
// Jump between ① ② ③ ④ ⑤
print(" Minimum Jump to end : ", task.minimumJumpsToEnd(arr2, n));
}
main();
Output
2 1 3 2 3 1 6 7 8 4 2 9
Minimum Jump to end : 4
2 3 0 1 2 3 0 1 2 6 4
Minimum Jump to end : 5
// Kotlin Program
// Program for minimum number of jumps to reach the end
class Jumping
{
// This is finding the number of minimum steps
// or jumps required to reach end of array elements
// When of array not contains negative values
fun minimumJumpsToEnd(arr: Array < Int > , n: Int): Int
{
if (n <= 1)
{
// When less than 2 elements
return 0;
}
if (arr[0] == 0)
{
// When first element is 0 then not possible to
// Visit next element
// Therefor
return -1;
}
// Use to collect number of jump process
var jump: Array < Int > = Array(n)
{
0
};
var i: Int = 1;
// Execute loop through by size of n
while (i < n)
{
// Set initial steps jump
jump[i] = Int.MAX_VALUE;
var j: Int = 0;
// Inner loop which is finding minimum jump between 0 to i
while (j < i)
{
if (i <= j + arr[j] && jump[j] != Int.MAX_VALUE)
{
if (jump[i] > (jump[j] + 1))
{
// Update new minimum jump
jump[i] = jump[j] + 1;
}
// Stop inner loop execution
j = i;
}
j += 1;
}
i += 1;
}
// When jump[ n-1] is MAX Integer
// That means we not reach the end of array elements
return jump[n - 1];
}
fun displayArray(arr: Array < Int > , n: Int): Unit
{
var i: Int = 0;
while (i < n)
{
print(" " + arr[i]);
i += 1;
}
print("\n");
}
}
fun main(args: Array < String > ): Unit
{
val task: Jumping = Jumping();
val arr1: Array < Int > = arrayOf(2, 1, 3, 2, 3, 1, 6, 7, 8, 4, 2, 9);
val arr2: Array < Int > = arrayOf(2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4);
// Test A
var n: Int = arr1.count();
task.displayArray(arr1, n);
// Given [2, 1, 3, 2, 3, 1 , 6, 7, 8, 4, 2, 9 ]
// Jump start with 2 → 3 → 1 → 6 → 9
// Jump between ① ② ③ ④
// Ans : 4
println(" Minimum Jump to end : " +
task.minimumJumpsToEnd(arr1, n));
// Test B
n = arr2.count();
task.displayArray(arr2, n);
// Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
// Jump start with 2 → 3 → 2 → 3 → 2 → 4
// Jump between ① ② ③ ④ ⑤
println(" Minimum Jump to end : " +
task.minimumJumpsToEnd(arr2, n));
}
Output
2 1 3 2 3 1 6 7 8 4 2 9
Minimum Jump to end : 4
2 3 0 1 2 3 0 1 2 6 4
Minimum Jump to end : 5
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