Review the fundamentals: Arrays guide and time complexities.

Maximized Partition Difference — Problem Statement & Solution Guide

Q: How can we precompute the necessary values to solve this in linear time?

We can use two auxiliary arrays. One array, `prefixMax`, where `prefixMax[i]` stores the maximum value in `nums[0...i]`. The other array, `suffixMin`, where `suffixMin[i]` stores the minimum value in `nums[i...n-1]`. By precomputing these in $O(n)$ time, we can evaluate the difference `prefixMax[k] - suffixMin[k+1]` for any $k$ in $O(1)$ time.

Q: Can this be solved with a single pass and constant extra space?

No, not directly in a single forward pass without storing the suffix minimums, because the suffix minimum depends on elements to the right of the current index. However, we can optimize space by computing the `suffixMin` array in $O(n)$ space and then iterating forward to compute the running `prefixMax` on the fly, reducing the space complexity to $O(n)$ instead of $O(2n)$, though $O(n)$ is still required for the suffix array. A true $O(1)$ space solution is not possible without re-reading the array or using a stack-like structure that doesn't apply here.

Q: What if the array is already sorted in ascending order? How does the answer change?

If the array is sorted in ascending order, the maximum of any prefix $nums[0...k]$ is always `nums[k]`, and the minimum of any suffix $nums[k+1...n-1]$ is always `nums[k+1]`. The difference becomes `nums[k] - nums[k+1]`, which will always be negative or zero (since `nums[k] <= nums[k+1]`). The maximum difference will be the least negative value, which occurs at the partition where the gap between consecutive elements is smallest. In the worst case, if all elements are equal, the difference is 0.

ArraysMediumBasic Traversal

TimeO(n)

SpaceO(n)

Quick Answer & Algorithm Key Takeaway

Iterating arrays and tracking min/max

TopicArrays

PatternBasic Traversal

TimeO(n)

SpaceO(n)

Problem Description

You are given an integer array nums of length n. You need to partition the array into two non-empty contiguous subarrays: a left subarray nums[0...k] and a right subarray nums[k+1...n-1], where 0 <= k < n - 1. Find the partition index k that maximizes the value of max(nums[0...k]) - min(nums[k+1...n-1]), and return this maximum difference.

DSA Pattern Breakdown

"Maximized Partition Difference"

medium

WHY DOES IT MATTER?

This problem exemplifies the 'Prefix-Suffix' or 'Precomputation' pattern, which is fundamental in optimizing problems that involve subarray properties. It teaches the critical skill of recognizing when redundant calculations can be eliminated by storing intermediate results, a technique widely used in dynamic programming, sliding window problems, and range query optimizations.

OPTIMIZATION CHALLENGE

The key insight is that the maximum of a prefix is non-decreasing as the prefix grows, and the minimum of a suffix is non-increasing as the suffix shrinks. This monotonicity allows us to compute these values in a single forward and backward pass, respectively, avoiding the $O(n^2)$ nested loops.

REAL-WORLD CONNECTION

Consider a stock trading platform where you need to find the best time to split a day's trading data into 'morning' and 'afternoon' sessions to maximize the difference between the highest morning price and the lowest afternoon price. Precomputing daily high/low aggregates allows real-time decision-making without re-scanning historical data for every potential split point.

In an interview, explicitly state the time complexity of the naive approach first to show you understand the problem's constraints. Then, propose the precomputation strategy, explaining that you are trading $O(n)$ space for $O(n)$ time. Be prepared to discuss if the space can be reduced, showing depth of understanding.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(n)

Core Theory — Why This Approach?

The problem requires finding a partition index $k$ that maximizes the difference between the maximum element in the prefix $nums[0...k]$ and the minimum element in the suffix $nums[k+1...n-1]$. A naive approach would iterate through every possible partition index $k$ and, for each, scan the left subarray to find its maximum and the right subarray to find its minimum. This results in a time complexity of $O(n^2)$, which is computationally expensive and fails for large input sizes where $n$ can be up to $10^5$ or more. The inefficiency stems from redundant calculations: the maximum of the prefix and the minimum of the suffix are recalculated from scratch for every single partition point, ignoring the overlapping nature of these subarrays.

Interview Questions on This Problem

Q1How can we precompute the necessary values to solve this in linear time?

We can use two auxiliary arrays. One array, prefixMax, where prefixMax[i] stores the maximum value in nums[0...i]. The other array, suffixMin, where suffixMin[i] stores the minimum value in nums[i...n-1]. By precomputing these in $O(n)$ time, we can evaluate the difference prefixMax[k] - suffixMin[k+1] for any $k$ in $O(1)$ time.

Q2Can this be solved with a single pass and constant extra space?

No, not directly in a single forward pass without storing the suffix minimums, because the suffix minimum depends on elements to the right of the current index. However, we can optimize space by computing the suffixMin array in $O(n)$ space and then iterating forward to compute the running prefixMax on the fly, reducing the space complexity to $O(n)$ instead of $O(2n)$, though $O(n)$ is still required for the suffix array. A true $O(1)$ space solution is not possible without re-reading the array or using a stack-like structure that doesn't apply here.

Q3What if the array is already sorted in ascending order? How does the answer change?

If the array is sorted in ascending order, the maximum of any prefix $nums[0...k]$ is always nums[k], and the minimum of any suffix $nums[k+1...n-1]$ is always nums[k+1]. The difference becomes nums[k] - nums[k+1], which will always be negative or zero (since nums[k] <= nums[k+1]). The maximum difference will be the least negative value, which occurs at the partition where the gap between consecutive elements is smallest. In the worst case, if all elements are equal, the difference is 0.

Examples

Example 1

Input

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

Output

Explanation: Partitioning at index k = 3 splits the array into [4, 3, 2, 5] (maximum is 5) and [1] (minimum is 1). This yields a difference of 5 - 1 = 4, which is the maximum achievable.

Example 2

Input

nums = [1, 10, 3, 2, 12]

Output

Explanation: Partitioning at index k = 1 splits the array into [1, 10] (maximum is 10) and [3, 2, 12] (minimum is 2). This yields a difference of 10 - 2 = 8, which is the maximum achievable.

Constraints

2 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9

Optimal Approach & Strategy

Precompute two arrays: prefixMax where prefixMax[i] is the max of nums[0...i], and suffixMin where suffixMin[i] is the min of nums[i...n-1]. Then, iterate through all valid k values, compute prefixMax[k] - suffixMin[k+1], and return the maximum difference found.

Brute Force Approach

Iterate through every possible partition index k from 0 to n-2. For each k, scan the subarray nums[0...k] to find its maximum and nums[k+1...n-1] to find its minimum, then compute the difference and update the global maximum.

Code Solutions

Language:

JavaScript Solution

Time: O(n)

function maximizedPartitionDifference(nums) {
    const n = nums.length;
    const leftMax = new Array(n);
    const rightMin = new Array(n);
    leftMax[0] = nums[0];
    for (let i = 1; i < n; i++) leftMax[i] = Math.max(leftMax[i - 1], nums[i]);
    rightMin[n - 1] = nums[n - 1];
    for (let i = n - 2; i >= 0; i--) rightMin[i] = Math.min(rightMin[i + 1], nums[i]);
    let maxDiff = -Infinity;
    for (let k = 0; k < n - 1; k++) {
        maxDiff = Math.max(maxDiff, leftMax[k] - rightMin[k + 1]);
    }
    return maxDiff;
}

#include <vector>
#include <algorithm>
#include <climits>
using namespace std;

class Solution {
public:
    long long maximizedPartitionDifference(vector<int>& nums) {
        int n = nums.size();
        if (n < 2) return 0;
        vector<long long> rightMin(n);
        rightMin[n-1] = nums[n-1];
        for (int i = n - 2; i >= 0; --i) {
            rightMin[i] = min(rightMin[i+1], (long long)nums[i]);
        }
        long long maxDiff = LLONG_MIN;
        long long leftMax = nums[0];
        for (int k = 0; k < n - 1; ++k) {
            leftMax = max(leftMax, (long long)nums[k]);
            maxDiff = max(maxDiff, leftMax - rightMin[k+1]);
        }
        return maxDiff;
    }
};

class Solution {
    public int maximizedPartitionDifference(int[] nums) {
        if (nums == null || nums.length < 2) {
            return 0;
        }
        int n = nums.length;
        int[] rightMin = new int[n];
        rightMin[n - 1] = nums[n - 1];
        for (int i = n - 2; i >= 0; i--) {
            rightMin[i] = Math.min(rightMin[i + 1], nums[i]);
        }
        
        long maxDiff = Long.MIN_VALUE;
        int leftMax = nums[0];
        for (int k = 0; k < n - 1; k++) {
            leftMax = Math.max(leftMax, nums[k]);
            long diff = (long) leftMax - (long) rightMin[k + 1];
            if (diff > maxDiff) {
                maxDiff = diff;
            }
        }
        return (int) maxDiff;
    }
}

from typing import List

def maximized_partition_difference(nums: List[int]) -> int:
    n = len(nums)
    if n < 2:
        return 0
    
    left_max = [0] * n
    right_min = [0] * n
    
    left_max[0] = nums[0]
    for i in range(1, n):
        left_max[i] = max(left_max[i-1], nums[i])
        
    right_min[n-1] = nums[n-1]
    for i in range(n-2, -1, -1):
        right_min[i] = min(right_min[i+1], nums[i])
        
    max_diff = left_max[0] - right_min[1]
    for k in range(1, n - 1):
        diff = left_max[k] - right_min[k+1]
        if diff > max_diff:
            max_diff = diff
            
    return max_diff

function maximizedPartitionDifference(nums) {
    const n = nums.length;
    const leftMax = new Array(n);
    const rightMin = new Array(n);
    leftMax[0] = nums[0];
    for (let i = 1; i < n; i++) leftMax[i] = Math.max(leftMax[i - 1], nums[i]);
    rightMin[n - 1] = nums[n - 1];
    for (let i = n - 2; i >= 0; i--) rightMin[i] = Math.min(rightMin[i + 1], nums[i]);
    let maxDiff = -Infinity;
    for (let k = 0; k < n - 1; k++) {
        maxDiff = Math.max(maxDiff, leftMax[k] - rightMin[k + 1]);
    }
    return maxDiff;
}

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