# 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
2 3 0 1 2 3 0 1 2 6 4
``````// 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;
// 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 : " +
// Test B
n = arr2.length;
// 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 : " +
}
}``````

#### Output

`````` 2 1 3 2 3 1 6 7 8 4 2 9
2 3 0 1 2 3 0 1 2 6 4
``````// 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]);
// 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]);
// 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
2 3 0 1 2 3 0 1 2 6 4
``````// 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;
// 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 : " +
// Test B
n = arr2.Length;
// 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 : " +
}
}``````

#### Output

`````` 2 1 3 2 3 1 6 7 8 4 2 9
2 3 0 1 2 3 0 1 2 6 4
``````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
2 3 0 1 2 3 0 1 2 6 4
``````<?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);
// 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 : ".
"\n");
// Test B
\$n = count(\$arr2);
// 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 : ".
"\n");
}
main();``````

#### Output

`````` 2 1 3 2 3 1 6 7 8 4 2 9
2 3 0 1 2 3 0 1 2 6 4
``````// 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;
// 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 : " +
// Test B
n = arr2.length;
// 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 : " +
}
main();``````

#### Output

`````` 2 1 3 2 3 1 6 7 8 4 2 9
2 3 0 1 2 3 0 1 2 6 4
``````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() :
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)
#  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 : ",
#  Test B
n = len(arr2)
#  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 : ",

if __name__ == "__main__": main()``````

#### Output

``````  2  1  3  2  3  1  6  7  8  4  2  9
2  3  0  1  2  3  0  1  2  6  4
``````#  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()
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
#  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 : ",
#  Test B
n = arr2.length
#  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 : ",
end

main()``````

#### Output

`````` 2 1 3 2 3 1 6 7 8 4 2 9
2 3 0 1 2 3 0 1 2 6 4
``````
``````// 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;
// 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 : " +
// Test B
n = arr2.length;
// 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 : " +
}
}``````

#### Output

`````` 2 1 3 2 3 1 6 7 8 4 2 9
2 3 0 1 2 3 0 1 2 6 4
``````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;
// 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
// Test B
n = arr2.count;
// 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
}
main();``````

#### Output

``````  2  1  3  2  3  1  6  7  8  4  2  9
2  3  0  1  2  3  0  1  2  6  4
``````// 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();
// 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 : " +
// Test B
n = arr2.count();
// 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 : " +
}``````

#### Output

`````` 2 1 3 2 3 1 6 7 8 4 2 9
2 3 0 1 2 3 0 1 2 6 4

## Comment

Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.