How do you solve Odd Even Jump on LeetCode?

Odd Even Jump (LeetCode #975) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Understanding the jump conditions is crucial.

What is the brute force solution for Odd Even Jump?

The brute force approach for Odd Even Jump is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible starting index and simulating the jumps until we either reach the end of the array or find that no further jumps are possible. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Odd Even Jump?

Odd Even Jump uses the following concepts: Array, Dynamic Programming, Stack, Sorting, Monotonic Stack, Ordered Set. The recommended patterns to study are: Dynamic Programming, Monotonic Stack, Sorting.

What are the interview tips for Odd Even Jump?

Clarify the problem statement before jumping into coding. Think about edge cases and how they affect your solution. Explain your thought process as you code to demonstrate your understanding.

What are common mistakes when solving Odd Even Jump?

Overlooking the conditions for odd and even jumps. Not considering edge cases like single-element arrays.

#975

Odd Even Jump

Hard

ArrayDynamic ProgrammingStackSortingMonotonic StackOrdered SetDynamic ProgrammingMonotonic StackSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution uses dynamic programming and a monotonic stack to efficiently determine which indices can reach the end of the array. This reduces the time complexity significantly by avoiding redundant checks.

⚙️

Algorithm

4 steps

1Step 1: Create two boolean arrays, oddJump and evenJump, initialized to false.
2Step 2: Set the last index of both arrays to true, as we can reach the end from itself.
3Step 3: Traverse the array backwards, using a monotonic stack to find valid jumps for odd and even jumps.
4Step 4: Update the oddJump and evenJump arrays based on the conditions for valid jumps.

solution.py25 lines

1def oddEvenJumps(arr):
2    n = len(arr)
3    if n == 1: return 1
4    oddJump = [False] * n
5    evenJump = [False] * n
6    oddJump[-1] = evenJump[-1] = True
7    oddCount = 0
8    
9    from sortedcontainers import SortedList
10    sortedList = SortedList()
11    sortedList.add((arr[-1], n - 1))
12    
13    for i in range(n - 2, -1, -1):
14        while sortedList and sortedList[0][0] < arr[i]:
15            sortedList.pop(0)
16        if sortedList:
17            oddJump[i] = evenJump[sortedList[0][1]]
18        
19        while sortedList and sortedList[-1][0] <= arr[i]:
20            sortedList.pop()
21        if sortedList:
22            evenJump[i] = oddJump[sortedList[-1][1]]
23        sortedList.add((arr[i], i))
24    
25    return sum(oddJump)

ℹ

Complexity note: The time complexity is O(n log n) because we use a sorted data structure to maintain the indices for the jumps, allowing us to efficiently find the next valid jump.

1Understanding the jump conditions is crucial.
2Using data structures like stacks can optimize the search for valid jumps.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.