What if you could know instantly if you can finish a tricky jump game without guessing every step?
Why Jump Game Problem in DSA Typescript?
Imagine you are playing a board game where each square tells you how many steps you can jump forward. You want to know if you can reach the last square starting from the first one. Doing this by checking every possible jump by hand is like trying to guess the right path without any clues.
Manually checking every jump is slow and confusing because you might try many wrong paths again and again. It's easy to get lost or make mistakes, especially if the board is long or has tricky jumps.
The Jump Game problem uses a smart way to check if you can reach the end by keeping track of the farthest place you can jump so far. This way, you don't waste time on impossible paths and quickly know if the last square is reachable.
function canReachEnd(jumps: number[]): boolean {
// Try every jump from each position manually
for (let i = 0; i < jumps.length; i++) {
for (let step = 1; step <= jumps[i]; step++) {
// Check if can jump to the end
}
}
return false;
}function canReachEnd(jumps: number[]): boolean {
let maxReach = 0;
for (let i = 0; i < jumps.length; i++) {
if (i > maxReach) return false;
maxReach = Math.max(maxReach, i + jumps[i]);
}
return true;
}This concept lets you quickly decide if you can reach the end without trying every path, saving time and effort.
Think of planning a road trip where each city sign tells you how far you can drive next. Using this method, you can quickly know if you can reach your destination without checking every possible route.
Manual checking is slow and confusing for jump paths.
Tracking the farthest reachable position speeds up the decision.
Helps solve reachability problems efficiently in games and planning.
Practice
Solution
Step 1: Understand the problem statement
The Jump Game problem asks if you can start at the first index and reach the last index by jumping forward using the numbers in the array as maximum jump lengths.Step 2: Identify the key goal
The goal is to check if the last index is reachable, not the number of jumps or sums.Final Answer:
If you can reach the last index starting from the first index by jumping according to the values. -> Option AQuick Check:
Jump Game = Reach last index? [OK]
- Confusing reachability with minimum jumps needed
- Thinking the problem asks for maximum jump length
- Assuming sum of jumps is the answer
Solution
Step 1: Analyze the input and output types
The input is an array of numbers representing jump lengths, so type is number[]. The output is a boolean indicating if last index is reachable.Step 2: Match the function signature
function canJump(nums: number[]): boolean matches input type number[] and output boolean, which fits the problem.Final Answer:
function canJump(nums: number[]): boolean -> Option AQuick Check:
Input array, output boolean = function canJump(nums: number[]): boolean [OK]
- Using wrong input types like string or 2D arrays
- Returning array instead of boolean
- Incorrect function parameter types
function canJump(nums: number[]): boolean {
let maxReach = 0;
for (let i = 0; i < nums.length; i++) {
if (i > maxReach) return false;
maxReach = Math.max(maxReach, i + nums[i]);
}
return true;
}
console.log(canJump([2,3,1,1,4]));Solution
Step 1: Trace the loop and maxReach updates
Start maxReach=0. i=0: maxReach=max(0,0+2)=2; i=1: 1<=2, maxReach=max(2,1+3)=4; i=2: 2<=4, maxReach=max(4,2+1)=4; i=3: 3<=4, maxReach=max(4,3+1)=4; i=4: 4<=4, maxReach=max(4,4+4)=8.Step 2: Check if loop completes without returning false
Since i never exceeds maxReach, loop completes and returns true.Final Answer:
true -> Option DQuick Check:
All indices reachable, so true [OK]
- Not updating maxReach correctly
- Returning false too early
- Confusing output with minimum jumps
function canJump(nums: number[]): boolean {
let maxReach = 0;
for (let i = 0; i <= nums.length; i++) {
if (i > maxReach) return false;
maxReach = Math.max(maxReach, i + nums[i]);
}
return true;
}Solution
Step 1: Check loop boundary conditions
The loop runs from i=0 to i <= nums.length, which means it accesses nums[nums.length] causing undefined access.Step 2: Correct the loop condition
Changing to i < nums.length prevents out-of-bounds access and fixes the bug.Final Answer:
The loop condition should be i < nums.length, not i <= nums.length. -> Option BQuick Check:
Loop index must stay within array bounds [OK]
- Using <= instead of < in loop condition
- Ignoring out-of-bounds array access
- Incorrect initialization of maxReach
[3,2,1,0,4], why does the standard greedy Jump Game algorithm return false?Solution
Step 1: Analyze jump reachability step-by-step
Start maxReach=0. i=0: maxReach=max(0,0+3)=3; i=1: 1<=3, maxReach=max(3,1+2)=3; i=2: 2<=3, maxReach=max(3,2+1)=3; i=3: 3<=3, maxReach=max(3,3+0)=3; i=4: 4>3, return false.Step 2: Understand why last index is unreachable
The zero at index 3 prevents extending maxReach beyond 3, so index 4 cannot be reached.Final Answer:
Because the zero at index 3 blocks any further jumps, making the last index unreachable. -> Option CQuick Check:
Zero jump blocks progress = false [OK]
- Ignoring zeros that block jumps
- Assuming array length affects result
- Misunderstanding maxReach updates
Ready to prove it?
Pass the test and this pattern moves unknown
