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Minimum Increase to Maximize Special Indices - Solution & Explanation

MediumArrayDynamic ProgrammingGreedyPrefix Sum12 min readAsked at: LinkedIn
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Problem Statement

You are given an integer array nums of length n.

An index i (0 < i < n - 1) is special if nums[i] > nums[i - 1] and nums[i] > nums[i + 1].

You may perform operations where you choose any index i and increase nums[i] by 1.

Your goal is to:

  • Maximize the number of special indices.
  • Minimize the total number of operations required to achieve that maximum.

Return an integer denoting the minimum total number of operations required.

 

Example 1:

Input: nums = [1,2,2]

Output: 1

Explanation:ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹

  • Start with nums = [1, 2, 2].
  • Increase nums[1] by 1, array becomes [1, 3, 2].
  • The final array is [1, 3, 2] has 1 special index, which is the maximum achievable.
  • It is impossible to achieve this number of special indices with fewer operations. Thus, the answer is 1.

Example 2:

Input: nums = [2,1,1,3]

Output: 2

Explanation:ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹

  • Start with nums = [2, 1, 1, 3].
  • Perform 2 operations at index 1, array becomes [2, 3, 1, 3].
  • The final array is [2, 3, 1, 3] has 1 special index, which is the maximum achievable. Thus, the answer is 2.

Example 3:

Input: nums = [5,2,1,4,3]

Output: 4

Explanation:ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹

  • Start with nums = [5, 2, 1, 4, 3].
  • Perform 4 operations at index 1, array becomes [5, 6, 1, 4, 3].
  • The final array is [5, 6, 1, 4, 3] has 2 special indices, which is the maximum achievable. Thus, the answer is 4.ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹

 

Constraints:

  • 3 <= n <= 105
  • 1 <= nums[i] <= 109

Approach Overview

Problem Overview: You are given an array and can increase elements by any amount. The goal is to apply the minimum total increase so the number of special indices becomes as large as possible. A special index is defined by a condition involving prefix and suffix values, which means the solution depends on efficiently tracking partial sums and evaluating each position.

Approach 1: Brute Force Simulation (O(n2) time, O(1) space)

The most direct strategy checks every index and determines how much increase is required to satisfy the special condition. For each position, recompute prefix sums and suffix sums and calculate the minimal increment needed to make the index valid. This leads to repeated scans of the array, producing quadratic time complexity. The approach is easy to reason about and useful for validating logic during development, but it becomes too slow for large inputs.

Approach 2: Prefix Sum Optimization (O(n) time, O(n) space)

Instead of recomputing sums repeatedly, precompute prefix sums and suffix sums once using a prefix sum array. With these values available, you can evaluate each index in constant time. The key insight is that the cost to make an index special can be expressed as a simple difference between prefix and suffix contributions. By iterating once through the array and computing the required increase for each position, you determine which indices can become special with minimal modification. This reduces the overall complexity to linear time.

Approach 3: Greedy with Running Prefix Tracking (O(n) time, O(1) space)

A further optimization avoids storing full auxiliary arrays. Maintain a running prefix sum while tracking the remaining suffix sum as you move through the array. For each index, compute the required increase using the current prefix state and the remaining suffix contribution. This greedy evaluation works because the cost for each index is independent once the prefix and suffix totals are known. The approach combines ideas from array traversal and prefix calculations to produce an optimal linear scan.

Recommended for interviews: Start by describing the brute force approach to show you understand the definition of a special index. Then move quickly to the prefix-sum based optimization. Interviewers typically expect the O(n) scan using prefix tracking because it demonstrates familiarity with prefix sum techniques and greedy reasoning. The optimal implementation is concise and scales well for large arrays.

Solution

We observe that if the array length is odd, then increasing all elements at odd indices so that each is 1 greater than both adjacent elements yields the maximum possible number of special indices. If the array length is even, then among indices in the range [1, n - 2], we skip exactly one index, and for the remaining indices, increase every other element so that each is 1 greater than both adjacent elements; this also yields the maximum possible number of special indices.

Therefore, we design a function dfs(i, j), which represents the minimum number of operations needed to obtain the maximum number of special indices starting from index i, with j remaining skips. For each index i, we can either increase it so that it is 1 greater than both neighbors, or skip it. We use memoized search to avoid repeated computation.

The implementation of dfs(i, j) is as follows:

  • If i geq n - 1, return 0.
  • Compute the number of operations required to increase nums[i] so that it is 1 greater than both adjacent elements, denoted as cost.
  • Compute the total cost for choosing to increase nums[i]: cost + dfs(i + 2, j).
  • If j > 0, compute the total cost for choosing to skip nums[i]: dfs(i + 1, 0), and update ans to the smaller of the two.

Finally, return dfs(1, (n bmod 2) \oplus 1).

The time complexity is O(n), and the space complexity is O(n), where n is the length of the array.

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Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Brute Force SimulationO(n^2)O(1)Useful for understanding the condition and verifying correctness on small inputs
Prefix Sum ArraysO(n)O(n)General solution when precomputing prefix and suffix sums simplifies evaluation
Greedy Prefix TrackingO(n)O(1)Optimal approach when memory usage should stay minimal

Video Solution

LeetCode Weekly Contest 496 - Q3 - Minimum Increase to Maximize Special Indices(3891) Explained ā€¢ Kumar K [Amazon] ā€¢ 1,159 views views

Watch 7 more video solutions ā†’

Frequently Asked Questions

Is Minimum Increase to Maximize Special Indices easy or hard?
The problem is generally classified as medium difficulty. The challenge comes from recognizing that repeated recomputation of prefix and suffix values is inefficient and that a prefix-sum based linear scan provides the optimal solution.
Minimum Increase to Maximize Special Indices Python/Java solution
Python and Java implementations typically iterate through the array once while maintaining prefix and suffix sums. Each index computes the required increment using simple arithmetic, giving O(n) time complexity and either O(n) or O(1) extra space.
How to solve Minimum Increase to Maximize Special Indices in O(n)?
Maintain a running prefix sum while also tracking the remaining suffix sum of the array. For each index, compute the minimal increase required to satisfy the special condition using these values. Because each step uses constant-time arithmetic, the entire algorithm completes in a single linear scan.
What is the best approach for Minimum Increase to Maximize Special Indices?
The optimal approach uses prefix sum calculations combined with a greedy scan of the array. Precompute or maintain running prefix and suffix sums so each index can be evaluated in constant time. This reduces the overall complexity to O(n) time while using either O(n) or O(1) additional space depending on implementation.
Is Minimum Increase to Maximize Special Indices asked at Google/Amazon/Meta?
Problems combining prefix sums, greedy reasoning, and array manipulation appear frequently in interviews at companies like Google, Amazon, and Meta. Variations that require computing prefix and suffix contributions efficiently are common in medium-difficulty interview rounds.
What data structure is used in Minimum Increase to Maximize Special Indices?
The main technique relies on arrays with prefix sum tracking. Some implementations store explicit prefix and suffix arrays, while optimized versions maintain running totals during a single pass.
What is the time complexity of Minimum Increase to Maximize Special Indices?
The optimized solution runs in O(n) time because each index is processed once while maintaining prefix and suffix information. A naive brute-force approach requires recomputing sums for each position and results in O(n^2) time.

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