Jump Game Problem in DSA C

Mental Model

You start at the first position and want to reach the last by jumping forward. Each number tells you the max steps you can jump from there.

Analogy: Imagine standing on stepping stones across a river. Each stone has a number showing how far you can jump forward. You want to see if you can reach the last stone without falling.

Index: 0   1   2   3   4
Array: [2] [3] [1] [1] [4]
Start ↑

Dry Run Walkthrough

Input: array: [2, 3, 1, 1, 4]

Goal: Determine if you can jump from start (index 0) to the last index (4)

Step 1: Start at index 0 with max jump 2, update furthest reachable to 2

furthest_reach = 2, current index = 0

Why: We can jump up to 2 steps from the first position

Step 2: Move to index 1, max jump 3, update furthest reachable to max(2,1+3)=4

furthest_reach = 4, current index = 1

Why: From index 1, we can jump further, so update the furthest reachable

Step 3: Move to index 2, max jump 1, furthest reachable stays 4

furthest_reach = 4, current index = 2

Why: No need to update since 2+1=3 < 4

Step 4: Move to index 3, max jump 1, furthest reachable stays 4

furthest_reach = 4, current index = 3

Why: No update needed, 3+1=4 equals current furthest

Step 5: Since furthest reachable (4) >= last index (4), return true

Reached last index

Why: We can reach or pass the last index

Result:

true -- can reach last index

Annotated Code

DSA C

#include <stdio.h>
#include <stdbool.h>

bool canJump(int* nums, int numsSize) {
    int furthest_reach = 0;
    for (int i = 0; i < numsSize; i++) {
        if (i > furthest_reach) {
            // Can't move forward anymore
            return false;
        }
        if (i + nums[i] > furthest_reach) {
            furthest_reach = i + nums[i];
        }
        if (furthest_reach >= numsSize - 1) {
            return true;
        }
    }
    return true;
}

int main() {
    int nums[] = {2, 3, 1, 1, 4};
    int size = sizeof(nums) / sizeof(nums[0]);
    bool result = canJump(nums, size);
    if (result) {
        printf("true\n");
    } else {
        printf("false\n");
    }
    return 0;
}

if (i > furthest_reach) { return false; }

stop if current index is beyond what we can reach

if (i + nums[i] > furthest_reach) { furthest_reach = i + nums[i]; }

update furthest reachable index if current jump extends it

if (furthest_reach >= numsSize - 1) { return true; }

return true early if we can reach or pass last index

Output

true

Complexity Analysis

Time: O(n) because we scan the array once updating the furthest reachable index

Space: O(1) because we use only a few variables regardless of input size

vs Alternative: Compared to recursive or backtracking approaches which can be exponential, this greedy method is much faster and uses less memory

Edge Cases

empty array

returns true because we are already at the last index

DSA C

for (int i = 0; i < numsSize; i++) { ... }

array with one element

returns true because start is last index

DSA C

if (furthest_reach >= numsSize - 1) { return true; }

array where first element is 0 and size > 1

returns false because can't move anywhere

DSA C

if (i > furthest_reach) { return false; }

When to Use This Pattern

When you see a problem asking if you can reach the end by jumping steps defined by array values, use the greedy furthest reachable index pattern to solve efficiently.

Common Mistakes

Mistake: Checking only if the current jump is possible without updating the furthest reachable index
Fix: Always update the furthest reachable index to track the maximum progress

Mistake: Not stopping early when the last index is reachable
Fix: Add a condition to return true as soon as furthest reachable index reaches or passes the last index

Summary

Determines if you can jump from start to end of array using max jump lengths at each position.

Use when you need to check reachability in a jump-based path problem.

Keep track of the furthest reachable index and stop early if you can reach the end.

Practice

1. What is the main goal of the Jump Game problem?

easy

A. Calculate the minimum jump length required to reach the last index

B. Determine if you can reach the last index from the first index by jumping

C. Count the total number of jumps needed to reach the end

D. Find the maximum jump length from the first index

Solution

1. Step 1: Understand the problem statement

   The Jump Game problem asks if it is possible to move from the first index to the last index by jumping forward using the values at each index as maximum jump lengths.

2. Step 2: Identify the goal

   The goal is to check if the last index is reachable, not to count jumps or find maximum jump length.

3. Final Answer:

   Determine if you can reach the last index from the first index by jumping -> Option B

4. Quick Check:

   Goal = Reach last index [OK]

Hint: Focus on reachability, not jump counts or lengths [OK]

Common Mistakes:

• Confusing maximum jump length with reachability
• Trying to count jumps instead of checking possibility
• Assuming jumps can go backward

2. Which of the following is the correct syntax to declare an integer array nums of size 5 in C for the Jump Game problem?

easy

A. int[] nums = new int[5];

B. int nums = [5];

C. int nums(5);

D. int nums[5];

Solution

1. Step 1: Recall C array declaration syntax

   In C, arrays are declared by specifying the type, name, and size in square brackets, like int nums[5];.

2. Step 2: Check each option

   int nums[5]; matches correct C syntax. int nums = [5]; uses invalid assignment. int nums(5); uses parentheses which is incorrect. int[] nums = new int[5]; is Java syntax.

3. Final Answer:

   int nums[5]; -> Option D

4. Quick Check:

   C array syntax = int nums[5]; [OK]

Hint: Use type name[size] for arrays in C [OK]

Common Mistakes:

• Using Java or C++ style array declarations
• Using parentheses instead of square brackets
• Assigning array size with equals sign

3. What will be the output of the following C code snippet for the Jump Game problem?

int nums[] = {2, 3, 1, 1, 4};
int n = 5;
int maxReach = 0;
for (int i = 0; i < n; i++) {
    if (i > maxReach) {
        printf("false\n");
        return 0;
    }
    if (i + nums[i] > maxReach) {
        maxReach = i + nums[i];
    }
}
printf("true\n");

medium

A. true

B. false

C. Compilation error

D. Runtime error

Solution

1. Step 1: Trace the loop and maxReach updates

   Start with maxReach=0. At i=0, maxReach = max(0, 0+2)=2. i=1 ≤ 2, maxReach = max(2,1+3)=4. i=2 ≤ 4, maxReach = max(4,2+1)=4. i=3 ≤ 4, maxReach = max(4,3+1)=4. i=4 ≤ 4, maxReach = max(4,4+4)=8.

2. Step 2: Check if loop completes without returning false

   Since i never exceeds maxReach, loop completes and prints "true".

3. Final Answer:

   true -> Option A

4. Quick Check:

   maxReach covers all indices = true [OK]

Hint: Check if current index exceeds max reachable index [OK]

Common Mistakes:

• Stopping early without updating maxReach
• Confusing index with jump length
• Assuming false if maxReach equals index

4. Identify the bug in the following C code snippet for the Jump Game problem:

int canJump(int* nums, int n) {
    int maxReach = 0;
    for (int i = 0; i <= n; i++) {
        if (i > maxReach) return 0;
        if (i + nums[i] > maxReach) maxReach = i + nums[i];
    }
    return 1;
}

medium

A. Return values should be swapped (1 for false, 0 for true)

B. maxReach should be initialized to -1

C. Loop condition should be i < n, not i <= n

D. nums array should be declared inside the function

Solution

1. Step 1: Analyze the loop condition

   The loop runs from i=0 to i<=n, which means it accesses nums[n], an out-of-bounds index causing undefined behavior.

2. Step 2: Correct the loop condition

   The loop should run while i < n to avoid accessing beyond the array.

3. Final Answer:

   Loop condition should be i < n, not i <= n -> Option C

4. Quick Check:

   Array index in bounds = i < n [OK]

Hint: Loop up to n-1 to avoid out-of-bounds access [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Ignoring array boundary checks
• Incorrect return values for boolean

5. Given the array nums = {3, 2, 1, 0, 4}, why does the standard greedy Jump Game algorithm return false for reachability?

hard

A. Because index 3 has a jump length of 0, blocking further progress

B. Because the array length is less than 5

C. Because the first element is too large

D. Because the last element is unreachable due to negative jumps

Solution

1. Step 1: Trace jumps from start

   Start at index 0 with jump length 3, maxReach = 3. Indices 1, 2, 3 are reachable. At index 3, jump length is 0, so maxReach stays 3.

2. Step 2: Check if last index is reachable

   Last index is 4, which is beyond maxReach 3, so it is unreachable, causing the algorithm to return false.

3. Final Answer:

   Because index 3 has a jump length of 0, blocking further progress -> Option A

4. Quick Check:

   Zero jump blocks path = false [OK]

Hint: Zero jump at reachable index blocks reaching the end [OK]

Common Mistakes:

• Ignoring zero jump blocking
• Assuming array length affects reachability
• Thinking negative jumps exist in problem