Global and Local Inversions

Global and Local Inversions - Problem

Array Math Medium

You are given an integer array nums of length n which represents a permutation of all the integers in the range [0, n - 1].

The number of global inversions is the number of the different pairs (i, j) where:

0 <= i < j < n
nums[i] > nums[j]

The number of local inversions is the number of indices i where:

0 <= i < n - 1
nums[i] > nums[i + 1]

Return true if the number of global inversions is equal to the number of local inversions.

Input & Output

Example 1 — Equal Inversions

$ Input: nums = [1,0,2]

› Output: true

💡 Note: Global inversions: (0,1) where 1 > 0. Local inversions: (0,1) where 1 > 0. Both counts are 1, so return true.

Example 2 — Different Inversions

$ Input: nums = [1,2,0]

› Output: false

💡 Note: Global inversions: (0,2) where 1 > 0 and (1,2) where 2 > 0 = 2 total. Local inversions: (1,2) where 2 > 0 = 1 total. 2 ≠ 1, so return false.

Example 3 — Already Sorted

$ Input: nums = [0,1,2,3]

› Output: true

💡 Note: No inversions of either type exist. 0 == 0, so return true.

Constraints

1 ≤ nums.length ≤ 10⁵
0 ≤ nums[i] < nums.length
nums is a permutation of [0, 1, 2, ..., nums.length - 1]

Visualization

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Asked in

G Google 15 f Facebook 12

The key insight is that every local inversion is also a global inversion. For the counts to be equal, there should be no global inversions that aren't local. This happens when each element is at most 1 position away from its sorted position. Best approach checks |nums[i] - i| ≤ 1 for all elements. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Count Both Types

⏱️ Time: O(n²) Space: O(1)

Count all global inversions by checking every pair (i,j) where i < j and nums[i] > nums[j]. Count local inversions by checking adjacent elements. Compare the counts.

Optimized - Check Position Constraint

⏱️ Time: O(n) Space: O(1)

Key insight: every local inversion is also a global inversion. For global == local, there should be no global inversions that aren't local. This happens when elements are at most 1 position away from their sorted position.

Brute Force - Count Both Types — Algorithm Steps

Count global inversions with nested loops
Count local inversions with single loop
Return true if counts are equal

Visualization

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Step-by-Step Walkthrough

Count Global

Check all pairs (i,j) where i < j

Count Local

Check adjacent pairs

Compare

Return if counts are equal

Code -

solution.c — C

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

bool solution(int* nums, int numsSize) {
    // Count global inversions
    int globalInv = 0;
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            if (nums[i] > nums[j]) {
                globalInv++;
            }
        }
    }
    
    // Count local inversions
    int localInv = 0;
    for (int i = 0; i < numsSize - 1; i++) {
        if (nums[i] > nums[i + 1]) {
            localInv++;
        }
    }
    
    return globalInv == localInv;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[1000];
    int size = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (token[0] >= '0' && token[0] <= '9') {
            nums[size++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    bool result = solution(nums, size);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops check all pairs for global inversions

⚠ Quadratic Growth

Space Complexity

O(1)

Only using counter variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Count Both Types — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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