LeetCode 3520 - Minimum Threshold for Inversion Pairs Count

Difficulty: 🟡 Medium
Topics: Array, Binary Search, Binary Indexed Tree, Segment Tree

Solution

Problem Understanding

The problem asks us to find the minimum threshold x such that there are at least k inversion pairs in an array nums of integers. An inversion pair (i, j) is defined by three conditions: first, the indices satisfy i < j; second, the values satisfy nums[i] > nums[j]; and third, the difference between the two values does not exceed the threshold x, i.e., nums[i] - nums[j] <= x. The input consists of an integer array nums of length up to 10⁴ with elements up to 10⁹, and an integer k up to 10⁹. The expected output is the smallest integer x such that there are at least k valid inversion pairs, or -1 if no such threshold exists.

The constraints indicate that a naive O(n²) solution will be too slow because n can be up to 10⁴, yielding up to 10⁸ comparisons. We must use a more efficient technique that leverages data structures supporting fast range counting, such as a Binary Indexed Tree (Fenwick Tree) or Segment Tree, combined with a binary search on the threshold x.

Important edge cases include arrays with all identical elements, arrays sorted in ascending or descending order, and scenarios where k is larger than the total possible inversion pairs.

Approaches

The brute-force approach examines all pairs (i, j) for i < j, counts those satisfying nums[i] > nums[j] and nums[i] - nums[j] <= x, and iterates through all possible thresholds to find the minimal one. While correct, this approach has time complexity O(n² log M) if we binary search over thresholds up to M = max(nums) - min(nums), which is too slow for n ≈ 10⁴.

The optimal approach leverages two key observations. First, the count of inversion pairs is non-decreasing with respect to the threshold x. Therefore, we can binary search on x to find the minimum threshold. Second, counting inversion pairs efficiently for a given x can be done using a Fenwick Tree or Segment Tree by iterating from right to left, mapping array values to ranks, and counting how many previously seen elements satisfy the threshold constraints.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n² log M)	O(1)	Checks all pairs for each possible threshold, too slow for n=10⁴
Optimal	O(n log n log M)	O(n)	Uses binary search on threshold and Fenwick Tree for counting inversion pairs

Algorithm Walkthrough

Identify the range of thresholds to search. Let lo = 0 and hi = max(nums) - min(nums). This ensures we cover all possible threshold differences.
Define a helper function count_pairs(threshold) which counts the number of inversion pairs for a given threshold.
Compress nums into ranks to make them suitable for Fenwick Tree indexing.
Initialize a Fenwick Tree of size equal to the number of unique ranks.
Iterate through nums from right to left. For each nums[i], query the tree for the number of elements already seen that satisfy nums[i] > nums[j] and nums[i] - nums[j] <= threshold.
Add the count to a running total and update the tree with the current element.
Perform a binary search on threshold x:
Compute mid = (lo + hi) // 2.
If count_pairs(mid) >= k, then mid is a candidate threshold, so set hi = mid.
Otherwise, set lo = mid + 1.
Continue until lo == hi.
If the final lo satisfies count_pairs(lo) >= k, return lo. Otherwise, return -1.

Why it works: The count of valid inversion pairs is monotonic with respect to the threshold. The binary search guarantees the minimum threshold is found, and the Fenwick Tree allows O(log n) counting per element, giving an efficient total runtime.

Python Solution

from typing import List
import bisect

class FenwickTree:
    def __init__(self, size: int):
        self.n = size
        self.tree = [0] * (self.n + 1)
    
    def update(self, index: int, delta: int):
        while index <= self.n:
            self.tree[index] += delta
            index += index & -index
    
    def query(self, index: int) -> int:
        res = 0
        while index > 0:
            res += self.tree[index]
            index -= index & -index
        return res

class Solution:
    def minThreshold(self, nums: List[int], k: int) -> int:
        if not nums or k <= 0:
            return -1

        sorted_unique = sorted(set(nums))
        def count_pairs(threshold: int) -> int:
            ft = FenwickTree(len(sorted_unique))
            total = 0
            for num in reversed(nums):
                left_val = num - threshold
                r = bisect.bisect_left(sorted_unique, left_val)
                total += ft.query(r)
                idx = bisect.bisect_left(sorted_unique, num) + 1
                ft.update(idx, 1)
            return total
        
        lo, hi = 0, max(nums) - min(nums)
        answer = -1
        while lo <= hi:
            mid = (lo + hi) // 2
            if count_pairs(mid) >= k:
                answer = mid
                hi = mid - 1
            else:
                lo = mid + 1
        return answer

The implementation defines a Fenwick Tree for prefix sum queries and rank compression for efficient indexing. The count_pairs function iterates from right to left, counting elements that satisfy the inversion condition with the current threshold. Binary search finds the minimal threshold satisfying the k inversion pairs requirement.

Go Solution

package main

import (
    "sort"
)

type FenwickTree struct {
    n int
    tree []int
}

func NewFenwickTree(size int) *FenwickTree {
    return &FenwickTree{n: size, tree: make([]int, size+1)}
}

func (ft *FenwickTree) Update(index, delta int) {
    for index <= ft.n {
        ft.tree[index] += delta
        index += index & -index
    }
}

func (ft *FenwickTree) Query(index int) int {
    res := 0
    for index > 0 {
        res += ft.tree[index]
        index -= index & -index
    }
    return res
}

func minThreshold(nums []int, k int) int {
    if len(nums) == 0 || k <= 0 {
        return -1
    }

    sortedUnique := append([]int(nil), nums...)
    sort.Ints(sortedUnique)
    sortedUnique = unique(sortedUnique)

    countPairs := func(threshold int) int {
        ft := NewFenwickTree(len(sortedUnique))
        total := 0
        for i := len(nums) - 1; i >= 0; i-- {
            leftVal := nums[i] - threshold
            r := sort.SearchInts(sortedUnique, leftVal)
            total += ft.Query(r)
            idx := sort.SearchInts(sortedUnique, nums[i]) + 1
            ft.Update(idx, 1)
        }
        return total
    }

    lo, hi := 0, max(nums) - min(nums)
    answer := -1
    for lo <= hi {
        mid := (lo + hi) / 2
        if countPairs(mid) >= k {
            answer = mid
            hi = mid - 1
        } else {
            lo = mid + 1
        }
    }
    return answer
}

func max(nums []int) int {
    m := nums[0]
    for _, v := range nums {
        if v > m {
            m = v
        }
    }
    return m
}

func min(nums []int) int {
    m := nums[0]
    for _, v := range nums {
        if v < m {
            m = v
        }
    }
    return m
}

func unique(nums []int) []int {
    res := []int{}
    for i, v := range nums {
        if i == 0 || v != nums[i-1] {
            res = append(res, v)
        }
    }
    return res
}

The Go implementation mirrors the Python version. Notable differences include handling slices versus lists, explicit helper functions for min, max, and unique values, and the lack of type hints. Fenwick Tree operations are identical in logic.

Worked Examples

Example 1: nums = [1,2,3,4,3,2,1], k = 7

Binary search starts with lo=0, hi=3. Count pairs for mid=1 yields fewer than 7, mid=2 yields ≥7, so the minimal threshold is 2. Fenwick Tree tracks counts of elements to the right for efficient querying.

Example 2: nums = [10,9,9,9,1], k = 4

Binary search over thresholds [0, 9] finds `mid