LeetCode #3578 — MEDIUM

Count Partitions With Max-Min Difference at Most K

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an integer array nums and an integer k. Your task is to partition nums into one or more non-empty contiguous segments such that in each segment, the difference between its maximum and minimum elements is at most k.

Return the total number of ways to partition nums under this condition.

Since the answer may be too large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [9,4,1,3,7], k = 4

Output: 6

Explanation:

There are 6 valid partitions where the difference between the maximum and minimum elements in each segment is at most k = 4:

[[9], [4], [1], [3], [7]]
[[9], [4], [1], [3, 7]]
[[9], [4], [1, 3], [7]]
[[9], [4, 1], [3], [7]]
[[9], [4, 1], [3, 7]]
[[9], [4, 1, 3], [7]]

Example 2:

Input: nums = [3,3,4], k = 0

Output: 2

Explanation:

There are 2 valid partitions that satisfy the given conditions:

[[3], [3], [4]]
[[3, 3], [4]]

Constraints:

2 <= nums.length <= 5 * 10⁴
1 <= nums[i] <= 10⁹
0 <= k <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and an integer k. Your task is to partition nums into one or more non-empty contiguous segments such that in each segment, the difference between its maximum and minimum elements is at most k. Return the total number of ways to partition nums under this condition. Since the answer may be too large, return it modulo 109 + 7.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Sliding Window · Monotonic Queue

Example 1

[9,4,1,3,7]
4

Example 2

[3,3,4]
0

Core Insight

What unlocks the optimal approach

Use dynamic programming.
Let <code>dp[idx]</code> be the count of ways to partition the array with the last partition ending at index <code>idx</code>.
Try using a sliding window; we can track the minimum and maximum in the window using deques.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3578: Count Partitions With Max-Min Difference at Most K
class Solution {
    public int countPartitions(int[] nums, int k) {
        final int mod = (int) 1e9 + 7;
        TreeMap<Integer, Integer> sl = new TreeMap<>();
        int n = nums.length;
        int[] f = new int[n + 1];
        int[] g = new int[n + 1];
        f[0] = 1;
        g[0] = 1;
        int l = 1;
        for (int r = 1; r <= n; r++) {
            int x = nums[r - 1];
            sl.merge(x, 1, Integer::sum);
            while (sl.lastKey() - sl.firstKey() > k) {
                if (sl.merge(nums[l - 1], -1, Integer::sum) == 0) {
                    sl.remove(nums[l - 1]);
                }
                ++l;
            }
            f[r] = (g[r - 1] - (l >= 2 ? g[l - 2] : 0) + mod) % mod;
            g[r] = (g[r - 1] + f[r]) % mod;
        }
        return f[n];
    }
}

// Accepted solution for LeetCode #3578: Count Partitions With Max-Min Difference at Most K
func countPartitions(nums []int, k int) int {
	const mod int = 1e9 + 7
	sl := redblacktree.New[int, int]()
	merge := func(st *redblacktree.Tree[int, int], x, v int) {
		c, _ := st.Get(x)
		if c+v == 0 {
			st.Remove(x)
		} else {
			st.Put(x, c+v)
		}
	}
	n := len(nums)
	f := make([]int, n+1)
	g := make([]int, n+1)
	f[0], g[0] = 1, 1
	for l, r := 1, 1; r <= n; r++ {
		merge(sl, nums[r-1], 1)
		for sl.Right().Key-sl.Left().Key > k {
			merge(sl, nums[l-1], -1)
			l++
		}
		f[r] = g[r-1]
		if l >= 2 {
			f[r] = (f[r] - g[l-2] + mod) % mod
		}
		g[r] = (g[r-1] + f[r]) % mod
	}
	return f[n]
}

# Accepted solution for LeetCode #3578: Count Partitions With Max-Min Difference at Most K
class Solution:
    def countPartitions(self, nums: List[int], k: int) -> int:
        mod = 10**9 + 7
        sl = SortedList()
        n = len(nums)
        f = [1] + [0] * n
        g = [1] + [0] * n
        l = 1
        for r, x in enumerate(nums, 1):
            sl.add(x)
            while sl[-1] - sl[0] > k:
                sl.remove(nums[l - 1])
                l += 1
            f[r] = (g[r - 1] - (g[l - 2] if l >= 2 else 0) + mod) % mod
            g[r] = (g[r - 1] + f[r]) % mod
        return f[n]

// Accepted solution for LeetCode #3578: Count Partitions With Max-Min Difference at Most K
use std::collections::BTreeMap;

impl Solution {
    pub fn count_partitions(nums: Vec<i32>, k: i32) -> i32 {
        const mod_val: i32 = 1_000_000_007;
        let n = nums.len();
        let mut f = vec![0; n + 1];
        let mut g = vec![0; n + 1];
        f[0] = 1;
        g[0] = 1;
        let mut sl = BTreeMap::new();
        let mut l = 1;
        for r in 1..=n {
            let x = nums[r - 1];
            *sl.entry(x).or_insert(0) += 1;
            while sl.keys().last().unwrap() - sl.keys().next().unwrap() > k {
                let val = nums[l - 1];
                if let Some(cnt) = sl.get_mut(&val) {
                    *cnt -= 1;
                    if *cnt == 0 {
                        sl.remove(&val);
                    }
                }
                l += 1;
            }
            f[r] = (g[r - 1] - if l >= 2 { g[l - 2] } else { 0 } + mod_val) % mod_val;
            g[r] = (g[r - 1] + f[r]) % mod_val;
        }
        f[n]
    }
}

// Accepted solution for LeetCode #3578: Count Partitions With Max-Min Difference at Most K
import { AvlTree } from 'datastructures-js';

function countPartitions(nums: number[], k: number): number {
    const mod = 1_000_000_007;
    const n = nums.length;

    const cnt = new Map<number, number>();
    const st = new AvlTree<number>((a, b) => a - b);

    function insert(x: number) {
        const c = (cnt.get(x) || 0) + 1;
        cnt.set(x, c);
        if (c === 1) {
            st.insert(x);
        }
    }

    function erase(x: number) {
        const c = cnt.get(x)! - 1;
        cnt.set(x, c);
        if (c === 0) {
            st.remove(x);
        }
    }

    const f = Array(n + 1).fill(0);
    const g = Array(n + 1).fill(0);
    f[0] = 1;
    g[0] = 1;

    for (let l = 1, r = 1; r <= n; ++r) {
        const x = nums[r - 1];
        insert(x);

        while (st.max()!.getValue() - st.min()!.getValue() > k) {
            erase(nums[l - 1]);
            l++;
        }

        f[r] = (g[r - 1] - (l >= 2 ? g[l - 2] : 0) + mod) % mod;
        g[r] = (g[r - 1] + f[r]) % mod;
    }

    return f[n];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.