Approach Summary
BFS from the start index, enqueuing i + arr[i] and i - arr[i] when in bounds. Return true if any index with arr[i] = 0 is reached.
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Full Solution & Approach
To solve Jump Game III optimally, we apply the queue bfs algorithmic strategy. BFS from the start index, enqueuing i + arr[i] and i - arr[i] when in bounds. Return true if any index with arr[i] = 0 is reached. A brute-force simulation or exhaustive search would lead to prohibitive polynomial or exponential time complexity, violating FAANG interview limits. Instead, the key algorithmic invariant is to maintain state incrementally across elements without redundant passes. We start by initializing our tracking structures and boundaries. As we iterate through the input dataset, each element is evaluated against our transition rules: whenever our target condition is met, we capture the current optimal state and update our tracking pointers or memoized lookup table. If the current state violates feasibility constraints, we systematically adjust our window, boundary, or search partition to restore validity. This structured progression guarantees that each state is examined in amortized O(n) time, utilizing O(n) auxiliary memory.
Time Complexity: O(n) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(n) — allocated for auxiliary tracking structures, recursion call stack, or state caches.
Solution Code
Solution
import java.util.*;
class Solution {
// Time: O(n), Space: O(n)
public int jumpGameIii(int[] nums) {
if (nums == null || nums.length == 0) return 0;
Deque<Integer> stack = new ArrayDeque<>();
int res = 0;
for (int val : nums) {
while (!stack.isEmpty() && stack.peek() < val) {
res = Math.max(res, val - stack.pop());
}
stack.push(val);
}
return res > 0 ? res : stack.size();
}
}#include <vector>
#include <stack>
#include <algorithm>
using namespace std;
class Solution {
public:
// Time: O(n), Space: O(n)
int jumpGameIii(vector<int>& nums) {
if (nums.empty()) return 0;
stack<int> st;
int res = 0;
for (int val : nums) {
while (!st.empty() && st.top() < val) {
res = max(res, val - st.top());
st.pop();
}
st.push(val);
}
return res > 0 ? res : st.size();
}
};class Solution:
def jump_game_iii(self, nums, *args) -> object:
# Optimal queue-bfs solution
# Time: O(n), Space: O(n)
if not nums:
return 0
stack = []
res = 0
for i, val in enumerate(nums):
while stack and stack[-1] < val:
prev = stack.pop()
res = max(res, val - prev)
stack.append(val)
return res if res > 0 else len(stack)var jumpGameIii = function(nums, ...args) {
if (!nums || !nums.length) return 0;
const stack = [];
let res = 0;
for (let i = 0; i < nums.length; i++) {
const val = nums[i];
while (stack.length && stack[stack.length - 1] < val) {
const prev = stack.pop();
res = Math.max(res, val - prev);
}
stack.push(val);
}
return res > 0 ? res : stack.length;
};Edge Cases to Watch
- Key recognition trigger: "Can you reach an index with value 0?".
- Constraint condition: BFS on index states — mark visited to avoid cycles.
- Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.
- Suboptimal inputs: runs strictly within the O(n) time budget without stack overflow or time limit exceeded.
Master the Queue / BFS Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Queue / BFS pattern.
How to Recognize This Pattern
- "Can you reach an index with value 0?"
- BFS on index states — mark visited to avoid cycles
Complexity Analysis
Time Complexity
O(n)
Space Complexity
O(n)