K Inverse Pairs Array - Practice Coding Problems

K Inverse Pairs Array - Problem

Dynamic Programming Hard

An inverse pair in an integer array nums is a pair of integers [i, j] where 0 <= i < j < nums.length and nums[i] > nums[j].

Given two integers n and k, return the number of different arrays consisting of numbers from 1 to n such that there are exactly k inverse pairs.

Since the answer can be huge, return it modulo 10^9 + 7.

Input & Output

Example 1 — Basic Case

$ Input: n = 3, k = 0

› Output: 1

💡 Note: Only one array [1,2,3] has 0 inverse pairs (already sorted)

Example 2 — One Inversion

$ Input: n = 3, k = 1

› Output: 2

💡 Note: Two arrays have exactly 1 inverse pair: [1,3,2] and [2,1,3]

Example 3 — Edge Case

$ Input: n = 4, k = 6

› Output: 1

💡 Note: Only [4,3,2,1] has maximum 6 inverse pairs for n=4

Constraints

1 ≤ n ≤ 1000
0 ≤ k ≤ 1000

Visualization

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Understanding the Visualization

Input

n=3 numbers [1,2,3], need k=1 inversions

Process

Find all arrangements with exactly 1 inversion

Output

Count = 2 valid arrangements

Key Takeaway

🎯 Key Insight: When placing the largest number at position p, it creates exactly (n-1-p) new inverse pairs with smaller numbers to its right

Asked in

G Google 25 M Microsoft 18 a Amazon 15

The key insight is that when placing the largest number in position p, it creates exactly (i-1-p) new inverse pairs. Best approach uses 2D DP with prefix sum optimization achieving Time: O(nk), Space: O(k).

Common Approaches

Approach	Time	Space	Notes
✓ Recursive with Memoization	O(n² × k²)	O(n × k)	Use top-down recursion with caching to avoid recomputation
Optimized DP with Prefix Sum	O(n × k)	O(k)	Optimize inner loop using prefix sum technique
Brute Force (Generate All Permutations)	O(n! × n²)	O(n)	Generate all possible arrays and count inversions for each
2D Dynamic Programming	O(n² × k)	O(n × k)	Build solution using bottom-up DP table

Recursive with Memoization — Algorithm Steps

Define dp(i, j) = number of ways to arrange first i numbers with j inverse pairs
Use recursion: place number i at position p creates (i-1-p) new inversions
Cache results to avoid recomputation

Visualization

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Step-by-Step Walkthrough

Base Case

dp(0,0) = 1, dp(0,j>0) = 0

Recursive

dp(i,j) sums over all valid positions

Memoize

Cache results to avoid recomputation

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

const int MOD = 1000000007;
int memo[1001][1001];
int memo_set[1001][1001];

int dp(int i, int j) {
    if (i == 0) {
        return j == 0 ? 1 : 0;
    }
    if (j < 0) return 0;
    
    if (memo_set[i][j]) {
        return memo[i][j];
    }
    
    int result = 0;
    int maxP = (i < j + 1) ? i : j + 1;
    
    for (int p = 0; p < maxP; p++) {
        result = (result + dp(i - 1, j - p)) % MOD;
    }
    
    memo[i][j] = result;
    memo_set[i][j] = 1;
    return result;
}

int solution(int n, int k) {
    memset(memo_set, 0, sizeof(memo_set));
    return dp(n, k);
}

int main() {
    int n, k;
    scanf("%d", &n);
    scanf("%d", &k);
    printf("%d\n", solution(n, k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n² × k²)

n×k states, each requiring O(n×k) computation without optimization

⚠ Quadratic Growth

Space Complexity

O(n × k)

Memoization table stores n×k states plus recursion stack

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Recursive with Memoization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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