Minimum number of jumps to reach end

• Difficulty Level : Medium
• Last Updated : 18 Oct, 2021

Given an array of integers where each element represents the max number of steps that can be made forward from that element. Write a function to return the minimum number of jumps to reach the end of the array (starting from the first element). If an element is 0, they cannot move through that element. If the end isn’t reachable, return -1.

Examples:

Input: arr[] = {1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9}
Output: 3 (1-> 3 -> 9 -> 9)
Explanation: Jump from 1st element
to 2nd element as there is only 1 step,
now there are three options 5, 8 or 9.
If 8 or 9 is chosen then the end node 9
can be reached. So 3 jumps are made.

Input:  arr[] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}
Output: 10
Explanation: In every step a jump
is needed so the count of jumps is 10.

The first element is 1, so can only go to 3. The second element is 3, so can make at most 3 steps eg to 5 or 8 or 9.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1: Naive Recursive Approach.
Approach: A naive approach is to start from the first element and recursively call for all the elements reachable from first element. The minimum number of jumps to reach end from first can be calculated using minimum number of jumps needed to reach end from the elements reachable from first.

minJumps(start, end) = Min ( minJumps(k, end) ) for all k reachable from start

C++

 // C++ program to find Minimum // number of jumps to reach end #include using namespace std;   // Function to return the minimum number // of jumps to reach arr[h] from arr[l] int minJumps(int arr[], int n) {       // Base case: when source and     // destination are same     if (n == 1)         return 0;       // Traverse through all the points     // reachable from arr[l]     // Recursively, get the minimum number     // of jumps needed to reach arr[h] from     // these reachable points     int res = INT_MAX;     for (int i = n - 2; i >= 0; i--) {         if (i + arr[i] >= n - 1) {             int sub_res = minJumps(arr, i + 1);             if (sub_res != INT_MAX)                 res = min(res, sub_res + 1);         }     }       return res; }   // Driver Code int main() {     int arr[] = { 1, 3, 6, 3, 2,                   3, 6, 8, 9, 5 };     int n = sizeof(arr) / sizeof(arr);     cout << "Minimum number of jumps to";     cout << " reach the end is " << minJumps(arr, n);     return 0; }   // This code is contributed // by Shivi_Aggarwal

C

 // C program to find Minimum // number of jumps to reach end #include #include   // Returns minimum number of // jumps to reach arr[h] from arr[l] int minJumps(int arr[], int l, int h) {     // Base case: when source and destination are same     if (h == l)         return 0;       // When nothing is reachable from the given source     if (arr[l] == 0)         return INT_MAX;       // Traverse through all the points     // reachable from arr[l]. Recursively     // get the minimum number of jumps     // needed to reach arr[h] from these     // reachable points.     int min = INT_MAX;     for (int i = l + 1; i <= h && i <= l + arr[l]; i++) {         int jumps = minJumps(arr, i, h);         if (jumps != INT_MAX && jumps + 1 < min)             min = jumps + 1;     }       return min; }   // Driver program to test above function int main() {     int arr[] = { 1, 3, 6, 3, 2, 3, 6, 8, 9, 5 };     int n = sizeof(arr) / sizeof(arr);     printf(         "Minimum number of jumps to reach end is %d ",         minJumps(arr, 0, n - 1));     return 0; }

Java

 // Java program to find Minimum // number of jumps to reach end import java.util.*; import java.io.*;   class GFG {     // Returns minimum number of     // jumps to reach arr[h] from arr[l]     static int minJumps(int arr[], int l, int h)     {         // Base case: when source         // and destination are same         if (h == l)             return 0;           // When nothing is reachable         // from the given source         if (arr[l] == 0)             return Integer.MAX_VALUE;           // Traverse through all the points         // reachable from arr[l]. Recursively         // get the minimum number of jumps         // needed to reach arr[h] from these         // reachable points.         int min = Integer.MAX_VALUE;         for (int i = l + 1; i <= h                             && i <= l + arr[l];              i++) {             int jumps = minJumps(arr, i, h);             if (jumps != Integer.MAX_VALUE && jumps + 1 < min)                 min = jumps + 1;         }         return min;     }       // Driver code     public static void main(String args[])     {         int arr[] = { 1, 3, 6, 3, 2, 3, 6, 8, 9, 5 };         int n = arr.length;         System.out.print("Minimum number of jumps to reach end is "                          + minJumps(arr, 0, n - 1));     } }   // This code is contributed by Sahil_Bansall

Python3

 # Python3 program to find Minimum # number of jumps to reach end   # Returns minimum number of jumps # to reach arr[h] from arr[l] def minJumps(arr, l, h):       # Base case: when source and     # destination are same     if (h == l):         return 0       # when nothing is reachable     # from the given source     if (arr[l] == 0):         return float('inf')       # Traverse through all the points     # reachable from arr[l]. Recursively     # get the minimum number of jumps     # needed to reach arr[h] from     # these reachable points.     min = float('inf')     for i in range(l + 1, h + 1):         if (i < l + arr[l] + 1):             jumps = minJumps(arr, i, h)             if (jumps != float('inf') and                        jumps + 1 < min):                 min = jumps + 1       return min   # Driver program to test above function arr = [1, 3, 6, 3, 2, 3, 6, 8, 9, 5] n = len(arr) print('Minimum number of jumps to reach',      'end is', minJumps(arr, 0, n-1))   # This code is contributed by Soumen Ghosh

C#

 // C# program to find Minimum // number of jumps to reach end using System;   class GFG {     // Returns minimum number of     // jumps to reach arr[h] from arr[l]     static int minJumps(int[] arr, int l, int h)     {         // Base case: when source         // and destination are same         if (h == l)             return 0;           // When nothing is reachable         // from the given source         if (arr[l] == 0)             return int.MaxValue;           // Traverse through all the points         // reachable from arr[l]. Recursively         // get the minimum number of jumps         // needed to reach arr[h] from these         // reachable points.         int min = int.MaxValue;         for (int i = l + 1; i <= h && i <= l + arr[l]; i++) {             int jumps = minJumps(arr, i, h);             if (jumps != int.MaxValue && jumps + 1 < min)                 min = jumps + 1;         }         return min;     }       // Driver code     public static void Main()     {         int[] arr = { 1, 3, 6, 3, 2, 3, 6, 8, 9, 5 };         int n = arr.Length;         Console.Write("Minimum number of jumps to reach end is "                       + minJumps(arr, 0, n - 1));     } }   // This code is contributed by Sam007



Javascript



Output:

Minimum number of jumps to reach end is 4

Complexity Analysis:

• Time complexity: O(n^n).
There are maximum n possible ways to move from a element. So maximum number of steps can be N^N so the upperbound of time complexity is O(n^n)
• Auxiliary Space: O(1).
There is no space required (if recursive stack space is ignored).

Note: If the execution is traced for this method, it can be seen that there will be overlapping subproblems. For example, minJumps(3, 9) will be called two times as arr is reachable from arr and arr. So this problem has both properties (optimal substructure and overlapping subproblems) of Dynamic Programming.

Method 2: Dynamic Programming.
Approach:

1. In this way, make a jumps[] array from left to right such that jumps[i] indicate the minimum number of jumps needed to reach arr[i] from arr.
2. To fill the jumps array run a nested loop inner loop counter is j and outer loop count is i.
3. Outer loop from 1 to n-1 and inner loop from 0 to i.
4. if i is less than j + arr[j] then set jumps[i] to minimum of jumps[i] and jumps[j] + 1. initially set jump[i] to INT MAX
5. Finally, return jumps[n-1].

C++

 // C++ program for Minimum number // of jumps to reach end #include using namespace std;   int min(int x, int y) { return (x < y) ? x : y; }   // Returns minimum number of jumps // to reach arr[n-1] from arr int minJumps(int arr[], int n) {     // jumps[n-1] will hold the result     int* jumps = new int[n];     int i, j;       if (n == 0 || arr == 0)         return INT_MAX;       jumps = 0;       // Find the minimum number of jumps to reach arr[i]     // from arr, and assign this value to jumps[i]     for (i = 1; i < n; i++) {         jumps[i] = INT_MAX;         for (j = 0; j < i; j++) {             if (i <= j + arr[j] && jumps[j] != INT_MAX) {                 jumps[i] = min(jumps[i], jumps[j] + 1);                 break;             }         }     }     return jumps[n - 1]; }   // Driver code int main() {     int arr[] = { 1, 3, 6, 1, 0, 9 };     int size = sizeof(arr) / sizeof(int);     cout << "Minimum number of jumps to reach end is "          << minJumps(arr, size);     return 0; }   // This is code is contributed by rathbhupendra

C

 // C program for Minimum number // of jumps to reach end #include #include   int min(int x, int y) { return (x < y) ? x : y; }   // Returns minimum number of // jumps to reach arr[n-1] from arr int minJumps(int arr[], int n) {     // jumps[n-1] will hold the result     int jumps[n];     int i, j;       if (n == 0 || arr == 0)         return INT_MAX;       jumps = 0;       // Find the minimum number of     // jumps to reach arr[i]     // from arr, and assign this     // value to jumps[i]     for (i = 1; i < n; i++) {         jumps[i] = INT_MAX;         for (j = 0; j < i; j++) {             if (i <= j + arr[j] && jumps[j] != INT_MAX) {                 jumps[i] = min(jumps[i], jumps[j] + 1);                 break;             }         }     }     return jumps[n - 1]; }   // Driver program to test above function int main() {     int arr[] = { 1, 3, 6, 1, 0, 9 };     int size = sizeof(arr) / sizeof(int);     printf("Minimum number of jumps to reach end is %d ",            minJumps(arr, size));     return 0; }

Java

 // JAVA Code for Minimum number // of jumps to reach end class GFG {       private static int minJumps(int[] arr, int n)     {         // jumps[n-1] will hold the         int jumps[] = new int[n];         // result         int i, j;           // if first element is 0,         if (n == 0 || arr == 0)             return Integer.MAX_VALUE;         // end cannot be reached           jumps = 0;           // Find the minimum number of jumps to reach arr[i]         // from arr, and assign this value to jumps[i]         for (i = 1; i < n; i++) {             jumps[i] = Integer.MAX_VALUE;             for (j = 0; j < i; j++) {                 if (i <= j + arr[j]                     && jumps[j]                            != Integer.MAX_VALUE) {                     jumps[i] = Math.min(jumps[i], jumps[j] + 1);                     break;                 }             }         }         return jumps[n - 1];     }       // driver program to test above function     public static void main(String[] args)     {         int arr[] = { 1, 3, 6, 1, 0, 9 };           System.out.println("Minimum number of jumps to reach end is : "                            + minJumps(arr, arr.length));     } }   // This code is contributed by Arnav Kr. Mandal.

Python3

 # Python3 program to find Minimum # number of jumps to reach end   # Returns minimum number of jumps # to reach arr[n-1] from arr def minJumps(arr, n):     jumps = [0 for i in range(n)]       if (n == 0) or (arr == 0):         return float('inf')       jumps = 0       # Find the minimum number of     # jumps to reach arr[i] from     # arr and assign this     # value to jumps[i]     for i in range(1, n):         jumps[i] = float('inf')         for j in range(i):             if (i <= j + arr[j]) and (jumps[j] != float('inf')):                 jumps[i] = min(jumps[i], jumps[j] + 1)                 break     return jumps[n-1]   # Driver Program to test above function arr = [1, 3, 6, 1, 0, 9] size = len(arr) print('Minimum number of jumps to reach',       'end is', minJumps(arr, size))   # This code is contributed by Soumen Ghosh

C#

 // C# Code for Minimum number of jumps to reach end using System;   class GFG {     static int minJumps(int[] arr, int n)     {         // jumps[n-1] will hold the         // result         int[] jumps = new int[n];           // if first element is 0,         if (n == 0 || arr == 0)               // end cannot be reached             return int.MaxValue;           jumps = 0;           // Find the minimum number of         // jumps to reach arr[i]         // from arr, and assign         // this value to jumps[i]         for (int i = 1; i < n; i++) {             jumps[i] = int.MaxValue;             for (int j = 0; j < i; j++) {                 if (i <= j + arr[j] && jumps[j] != int.MaxValue) {                     jumps[i] = Math.Min(jumps[i], jumps[j] + 1);                     break;                 }             }         }         return jumps[n - 1];     }       // Driver program     public static void Main()     {         int[] arr = { 1, 3, 6, 1, 0, 9 };         Console.Write("Minimum number of jumps to reach end is : " + minJumps(arr, arr.Length));     } }   // This code is contributed by Sam007



Javascript



Output:

Minimum number of jumps to reach end is 3

Thanks to paras for suggesting this method.
Time Complexity: O(n^2)

Auxiliary Space: O(n)

Method 3: Dynamic Programming.
In this method, we build jumps[] array from right to left such that jumps[i] indicates the minimum number of jumps needed to reach arr[n-1] from arr[i]. Finally, we return jumps.

C++

 // C++ program to find Minimum // number of jumps to reach end #include using namespace std;   // Returns Minimum number of // jumps to reach end int minJumps(int arr[], int n) {     // jumps will hold the result     int* jumps = new int[n];     int min;       // Minimum number of jumps needed     // to reach last element from last     // elements itself is always 0     jumps[n - 1] = 0;       // Start from the second element,     // move from right to left and     // construct the jumps[] array where     // jumps[i] represents minimum number     // of jumps needed to reach     // arr[m-1] from arr[i]     for (int i = n - 2; i >= 0; i--) {         // If arr[i] is 0 then arr[n-1]         // can't be reached from here         if (arr[i] == 0)             jumps[i] = INT_MAX;           // If we can directly reach to         // the end point from here then         // jumps[i] is 1         else if (arr[i] >= n - i - 1)             jumps[i] = 1;           // Otherwise, to find out the minimum         // number of jumps needed to reach         // arr[n-1], check all the points         // reachable from here and jumps[]         // value for those points         else {             // initialize min value             min = INT_MAX;               // following loop checks with all             // reachable points and takes             // the minimum             for (int j = i + 1; j < n && j <= arr[i] + i; j++) {                 if (min > jumps[j])                     min = jumps[j];             }               // Handle overflow             if (min != INT_MAX)                 jumps[i] = min + 1;             else                 jumps[i] = min; // or INT_MAX         }     }       return jumps; }   // Driver program to test above function int main() {     int arr[] = { 1, 3, 6, 1, 0, 9 };     int size = sizeof(arr) / sizeof(int);     cout << "Minimum number of jumps to reach"          << " end is " << minJumps(arr, size);     return 0; }

Java

 // Java program to find Minimum // number of jumps to reach end class GFG {     // Returns Minimum number     // of jumps to reach end     static int minJumps(int arr[],                         int n)     {         // jumps will         // hold the result         int[] jumps = new int[n];         int min;           // Minimum number of jumps         // needed to reach last         // element from last elements         // itself is always 0         jumps[n - 1] = 0;           // Start from the second         // element, move from right         // to left and construct the         // jumps[] array where jumps[i]         // represents minimum number of         // jumps needed to reach arr[m-1]         // from arr[i]         for (int i = n - 2; i >= 0; i--) {             // If arr[i] is 0 then arr[n-1]             // can't be reached from here             if (arr[i] == 0)                 jumps[i] = Integer.MAX_VALUE;               // If we can directly reach to             // the end point from here then             // jumps[i] is 1             else if (arr[i] >= n - i - 1)                 jumps[i] = 1;               // Otherwise, to find out             // the minimum number of             // jumps needed to reach             // arr[n-1], check all the             // points reachable from             // here and jumps[] value             // for those points             else {                 // initialize min value                 min = Integer.MAX_VALUE;                   // following loop checks                 // with all reachable points                 // and takes the minimum                 for (int j = i + 1; j < n && j <= arr[i] + i; j++) {                     if (min > jumps[j])                         min = jumps[j];                 }                   // Handle overflow                 if (min != Integer.MAX_VALUE)                     jumps[i] = min + 1;                 else                     jumps[i] = min; // or Integer.MAX_VALUE             }         }           return jumps;     }       // Driver Code     public static void main(String[] args)     {         int[] arr = { 1, 3, 6, 1, 0, 9 };         int size = arr.length;         System.out.println("Minimum number of"                            + " jumps to reach end is " + minJumps(arr, size));     } }   // This code is contributed by mits.

Python3

 # Python3 program to find Minimum # number of jumps to reach end   # Returns Minimum number of # jumps to reach end def minJumps(arr, n):           # jumps will hold the result     jumps = [0 for i in range(n)]       # Minimum number of jumps needed     # to reach last element from     # last elements itself is always 0     # jumps[n-1] is also initialized to 0       # Start from the second element,     # move from right to left and     # construct the jumps[] array where     # jumps[i] represents minimum number     # of jumps needed to reach arr[m-1]     # form arr[i]     for i in range(n-2, -1, -1):                   # If arr[i] is 0 then arr[n-1]         # can't be reached from here         if (arr[i] == 0):             jumps[i] = float('inf')           # If we can directly reach to         # the end point from here then         # jumps[i] is 1         elif (arr[i] >= n - i - 1):             jumps[i] = 1           # Otherwise, to find out the         # minimum number of jumps         # needed to reach arr[n-1],         # check all the points         # reachable from here and         # jumps[] value for those points         else:             # initialize min value             min = float('inf')               # following loop checks with             # all reachable points and             # takes the minimum             for j in range(i + 1, n):                 if (j <= arr[i] + i):                     if (min > jumps[j]):                         min = jumps[j]                                       # Handle overflow             if (min != float('inf')):                 jumps[i] = min + 1             else:                 # or INT_MAX                 jumps[i] = min       return jumps   # Driver program to test above function arr = [1, 3, 6, 3, 2, 3, 6, 8, 9, 5] n = len(arr) print('Minimum number of jumps to reach',       'end is', minJumps(arr, n-1))         # This code is contributed by Soumen Ghosh

C#

 // C# program to find Minimum // number of jumps to reach end using System;   class GFG {     // Returns Minimum number     // of jumps to reach end     public static int minJumps(int[] arr, int n)     {         // jumps will         // hold the result         int[] jumps = new int[n];         int min;           // Minimum number of jumps needed to         // reach last element from last elements         // itself is always 0         jumps[n - 1] = 0;           // Start from the second element, move         // from right to left and construct the         // jumps[] array where jumps[i] represents         // minimum number of jumps needed to reach         // arr[m-1] from arr[i]         for (int i = n - 2; i >= 0; i--) {             // If arr[i] is 0 then arr[n-1]             // can't be reached from here             if (arr[i] == 0) {                 jumps[i] = int.MaxValue;             }               // If we can directly reach to the end             // point from here then jumps[i] is 1             else if (arr[i] >= n - i - 1) {                 jumps[i] = 1;             }               // Otherwise, to find out the minimum             // number of jumps needed to reach             // arr[n-1], check all the points             // reachable from here and jumps[] value             // for those points             else {                 // initialize min value                 min = int.MaxValue;                   // following loop checks with all                 // reachable points and takes the minimum                 for (int j = i + 1; j < n && j <= arr[i] + i; j++) {                     if (min > jumps[j]) {                         min = jumps[j];                     }                 }                   // Handle overflow                 if (min != int.MaxValue) {                     jumps[i] = min + 1;                 }                 else {                     jumps[i] = min; // or Integer.MAX_VALUE                 }             }         }           return jumps;     }       // Driver Code     public static void Main(string[] args)     {         int[] arr = new int[] { 1, 3, 6, 1, 0, 9 };         int size = arr.Length;         Console.WriteLine("Minimum number of"                           + " jumps to reach end is " + minJumps(arr, size));     } }   // This code is contributed by Shrikant13

PHP

 = 0; \$i--)     {         // If arr[i] is 0 then arr[n-1]         // can't be reached from here         if (\$arr[\$i] == 0)             \$jumps[\$i] = PHP_INT_MAX;           // If we can directly reach to         // the end point from here then         // jumps[i] is 1         else if (\$arr[\$i] >= (\$n - \$i) - 1)             \$jumps[\$i] = 1;           // Otherwise, to find out the minimum         // number of jumps needed to reach         // arr[n-1], check all the points         // reachable from here and jumps[]         // value for those points         else         {             // initialize min value             \$min = PHP_INT_MAX;               // following loop checks with all             // reachable points and takes             // the minimum             for (\$j = \$i + 1; \$j < \$n &&                  \$j <= \$arr[\$i] + \$i; \$j++)             {                 if (\$min > \$jumps[\$j])                     \$min = \$jumps[\$j];             }               // Handle overflow             if (\$min != PHP_INT_MAX)                 \$jumps[\$i] = \$min + 1;             else                 \$jumps[\$i] = \$min; // or INT_MAX         }     }       return \$jumps; }   // Driver Code \$arr = array(1, 3, 6, 1, 0, 9); \$size = sizeof(\$arr); echo "Minimum number of jumps to reach",      " end is ", minJumps(\$arr, \$size);   // This code is contributed by ajit. ?>

Javascript



Output:

Minimum number of jumps to reach end is 3

Complexity Analysis:

• Time complexity:O(n^2).
Nested traversal of the array is needed.
• Auxiliary Space:O(n).
To store the DP array linear space is needed.

Minimum number of jumps to reach end | Set 2 (O(n) solution)
Thanks to Ashish for suggesting this solution.