Jump Game II
A medium Dynamic Programming problem included in Love Babbar 450, Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Dynamic Programming
- Sheets
- 2
- Core for
- 9 roles
- Platform
- LeetCode
The problem
Given an array of non-negative integers where each element represents your maximum jump length at that position, find the minimum number of jumps needed to reach the last index.
Example 1
- Input
- nums = [2,3,1,1,4]
- Output
- 2
- Why
- The minimum number of jumps is 2: jump 1 step from index 0 to 1, then 3 steps to the last index.
Example 2
- Input
- nums = [2,3,0,1,4]
- Output
- 2
- Why
- Jump from index 0 to 1, then from index 1 jump 3 steps to the end.
Example 3
- Input
- nums = [1]
- Output
- 0
- Why
- Already at the last index, no jumps needed.
Constraints
- 1 <= nums.length <= 10^4
- 0 <= nums[i] <= 1000
- It is guaranteed you can reach the last index
How to think about it
Updated 2026-09-09Think of jumps as widening horizons or breadth-first search levels on an array. With each jump, you can reach anywhere up to the current level's boundary; while stepping through that window, you scout the furthest landing spot for the next jump.
Approaches, worst first
Backtracking minimum jumps
time O(2^n) · space O(n)
Explore every valid jump length from each position and take the minimum across all successful paths to the end. Re-evaluates identical suffixes countless times.
Dynamic programming tabulation
time O(n^2) · space O(n)
Let dp[i] be the minimum jumps to reach index i. Initialize dp with infinity, set dp[0] = 0, and for each index i, update dp[j] = min(dp[j], dp[i] + 1) for all reachable j.
Greedy BFS interval expansionWrite this one
time O(n) · space O(1)
Maintain `currentEnd` (end of current jump's reach) and `farthest`. Iterate i from 0 to n - 2: update farthest = max(farthest, i + nums[i]). When i reaches currentEnd, increment jumps and set currentEnd = farthest.
Where people lose marks · 3
- Iterating the loop all the way to n - 1. If i reaches n - 1 and equals currentEnd, it triggers an unnecessary extra jump count when you are already at the destination.
- Input with n = 1 requires 0 jumps because no movement is necessary; iterating incorrectly may report 1.
- Forgetting to update farthest at every index within the current jump range, which artificially shrinks the reach of the next step.
The theory behind it
Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems
What Dynamic Programming is
Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.
When to reach for it
Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.
How the pattern works
Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.
What each operation costs
| Operation | Time |
|---|---|
| fill dynamic programming table of n states | O(n) |
| solve two-dimensional grid of m by n states | O(m * n) |
| space-optimized state transition keeping one row | O(n) |
What usually goes wrong with Dynamic Programming
- Filling a bottom-up table in an order where the current cell needs values that have not been computed yet, reading uninitialized zeros.
- Failing to initialize base cases properly, such as filling a minimization table with zeros instead of infinity, which traps the answer at zero.
- Overwriting values in a 1D space-optimized knapsack array by scanning in the wrong direction, allowing the same item to be chosen multiple times.
Which roles need this problem
Dynamic Programming is a core topic for these 9 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 5 more roles, including Game Developer, Cryptography Engineer, Performance Engineer.
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