LeetCode #3193 — HARD

Count the Number of Inversions

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an integer n and a 2D array requirements, where requirements[i] = [end_i, cnt_i] represents the end index and the inversion count of each requirement.

A pair of indices (i, j) from an integer array nums is called an inversion if:

i < j and nums[i] > nums[j]

Return the number of permutations perm of [0, 1, 2, ..., n - 1] such that for all requirements[i], perm[0..end_i] has exactly cnt_i inversions.

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

Example 1:

Input: n = 3, requirements = [[2,2],[0,0]]

Output: 2

Explanation:

The two permutations are:

[2, 0, 1]
- Prefix [2, 0, 1] has inversions (0, 1) and (0, 2).
- Prefix [2] has 0 inversions.
[1, 2, 0]
- Prefix [1, 2, 0] has inversions (0, 2) and (1, 2).
- Prefix [1] has 0 inversions.

Example 2:

Input: n = 3, requirements = [[2,2],[1,1],[0,0]]

Output: 1

Explanation:

The only satisfying permutation is [2, 0, 1]:

Prefix [2, 0, 1] has inversions (0, 1) and (0, 2).
Prefix [2, 0] has an inversion (0, 1).
Prefix [2] has 0 inversions.

Example 3:

Input: n = 2, requirements = [[0,0],[1,0]]

Output: 1

Explanation:

The only satisfying permutation is [0, 1]:

Prefix [0] has 0 inversions.
Prefix [0, 1] has no inversions.

Constraints:

2 <= n <= 300
1 <= requirements.length <= n
requirements[i] = [end_i, cnt_i]
0 <= end_i <= n - 1
0 <= cnt_i <= 400
The input is generated such that there is at least one i such that end_i == n - 1.
The input is generated such that all end_i are unique.

Step 01

Brute Force Baseline

Problem summary: You are given an integer n and a 2D array requirements, where requirements[i] = [endi, cnti] represents the end index and the inversion count of each requirement. A pair of indices (i, j) from an integer array nums is called an inversion if: i < j and nums[i] > nums[j] Return the number of permutations perm of [0, 1, 2, ..., n - 1] such that for all requirements[i], perm[0..endi] has exactly cnti inversions. Since the answer may be very 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

Example 1

3
[[2,2],[0,0]]

Example 2

3
[[2,2],[1,1],[0,0]]

Example 3

2
[[0,0],[1,0]]

Core Insight

What unlocks the optimal approach

Let <code>dp[i][j]</code> denote the number of arrays of length <code>i</code> with <code>j</code> inversions.
<code>dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1] + … + dp[i - 1][0]</code>.
<code>dp[i][j] = 0</code> if for some <code>x</code>, <code>requirements[x][0] == i</code> and <code>requirements[x][1] != j</code>.

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

Largest constraint values

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 #3193: Count the Number of Inversions
class Solution {
    public int numberOfPermutations(int n, int[][] requirements) {
        int[] req = new int[n];
        Arrays.fill(req, -1);
        int m = 0;
        for (var r : requirements) {
            req[r[0]] = r[1];
            m = Math.max(m, r[1]);
        }
        if (req[0] > 0) {
            return 0;
        }
        req[0] = 0;
        final int mod = (int) 1e9 + 7;
        int[][] f = new int[n][m + 1];
        f[0][0] = 1;
        for (int i = 1; i < n; ++i) {
            int l = 0, r = m;
            if (req[i] >= 0) {
                l = r = req[i];
            }
            for (int j = l; j <= r; ++j) {
                for (int k = 0; k <= Math.min(i, j); ++k) {
                    f[i][j] = (f[i][j] + f[i - 1][j - k]) % mod;
                }
            }
        }
        return f[n - 1][req[n - 1]];
    }
}

// Accepted solution for LeetCode #3193: Count the Number of Inversions
func numberOfPermutations(n int, requirements [][]int) int {
	req := make([]int, n)
	for i := range req {
		req[i] = -1
	}
	for _, r := range requirements {
		req[r[0]] = r[1]
	}
	if req[0] > 0 {
		return 0
	}
	req[0] = 0
	m := slices.Max(req)
	const mod = int(1e9 + 7)
	f := make([][]int, n)
	for i := range f {
		f[i] = make([]int, m+1)
	}
	f[0][0] = 1
	for i := 1; i < n; i++ {
		l, r := 0, m
		if req[i] >= 0 {
			l, r = req[i], req[i]
		}
		for j := l; j <= r; j++ {
			for k := 0; k <= min(i, j); k++ {
				f[i][j] = (f[i][j] + f[i-1][j-k]) % mod
			}
		}
	}
	return f[n-1][req[n-1]]
}

# Accepted solution for LeetCode #3193: Count the Number of Inversions
class Solution:
    def numberOfPermutations(self, n: int, requirements: List[List[int]]) -> int:
        req = [-1] * n
        for end, cnt in requirements:
            req[end] = cnt
        if req[0] > 0:
            return 0
        req[0] = 0
        mod = 10**9 + 7
        m = max(req)
        f = [[0] * (m + 1) for _ in range(n)]
        f[0][0] = 1
        for i in range(1, n):
            l, r = 0, m
            if req[i] >= 0:
                l = r = req[i]
            for j in range(l, r + 1):
                for k in range(min(i, j) + 1):
                    f[i][j] = (f[i][j] + f[i - 1][j - k]) % mod
        return f[n - 1][req[n - 1]]

// Accepted solution for LeetCode #3193: Count the Number of Inversions
/**
 * [3193] Count the Number of Inversions
 */
pub struct Solution {}

// submission codes start here
use std::collections::HashMap;

const MOD: i32 = 1_000_000_007;

impl Solution {
    pub fn number_of_permutations(n: i32, requirements: Vec<Vec<i32>>) -> i32 {
        let n = n as usize;
        let mut requirements_map = HashMap::new();
        let mut max_pair_count = 0;

        requirements_map.insert(0, 0);
        for requirement in requirements {
            requirements_map.insert(requirement[0] as usize, requirement[1] as usize);
            max_pair_count = max_pair_count.max(requirement[1]);
        }

        if let Some(count) = requirements_map.get(&0) {
            if *count != 0 {
                return 0;
            }
        }

        let mut dp = vec![vec![-1; max_pair_count as usize + 1]; n];

        if let Some(&count) = requirements_map.get(&(n - 1)) {
            Self::dfs(n - 1, count, &requirements_map, &mut dp)
        } else {
            Self::dfs(n - 1, 0, &requirements_map, &mut dp)
        }
    }

    fn dfs(
        end: usize,
        count: usize,
        requirements: &HashMap<usize, usize>,
        dp: &mut Vec<Vec<i32>>,
    ) -> i32 {
        if end == 0 {
            return 1;
        }

        if dp[end][count] != -1 {
            return dp[end][count];
        }

        if let Some(&c) = requirements.get(&(end - 1)) {
            if c <= count && count <= end + c {
                let r = Self::dfs(end - 1, c, requirements, dp);
                dp[end][count] = r;
                r
            } else {
                dp[end][count] = 0;
                0
            }
        } else {
            let mut r = 0;
            for i in 0..=end.min(count) {
                r = (r + Self::dfs(end - 1, count - i, requirements, dp)) % MOD;
            }

            dp[end][count] = r;
            r
        }
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3193() {
        assert_eq!(1, Solution::number_of_permutations(2, vec![vec![1, 1]]));
        assert_eq!(
            2,
            Solution::number_of_permutations(3, vec![vec![2, 2], vec![0, 0]])
        );
        assert_eq!(
            1,
            Solution::number_of_permutations(3, vec![vec![2, 2], vec![1, 1], vec![0, 0]])
        );
        assert_eq!(
            1,
            Solution::number_of_permutations(2, vec![vec![0, 0], vec![1, 0]])
        );
    }
}

// Accepted solution for LeetCode #3193: Count the Number of Inversions
function numberOfPermutations(n: number, requirements: number[][]): number {
    const req: number[] = Array(n).fill(-1);
    for (const [end, cnt] of requirements) {
        req[end] = cnt;
    }
    if (req[0] > 0) {
        return 0;
    }
    req[0] = 0;
    const m = Math.max(...req);
    const mod = 1e9 + 7;
    const f = Array.from({ length: n }, () => Array(m + 1).fill(0));
    f[0][0] = 1;
    for (let i = 1; i < n; ++i) {
        let [l, r] = [0, m];
        if (req[i] >= 0) {
            l = r = req[i];
        }
        for (let j = l; j <= r; ++j) {
            for (let k = 0; k <= Math.min(i, j); ++k) {
                f[i][j] = (f[i][j] + f[i - 1][j - k]) % mod;
            }
        }
    }
    return f[n - 1][req[n - 1]];
}

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 × m × \min(n, m)

Space

O(n × m)

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.

Count the Number of Inversions

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Off-by-one on range boundaries

State misses one required dimension