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.
Move from brute-force thinking to an efficient approach using array strategy.
You are given a list of preferences for n friends, where n is always even.
For each person i, preferences[i] contains a list of friends sorted in the order of preference. In other words, a friend earlier in the list is more preferred than a friend later in the list. Friends in each list are denoted by integers from 0 to n-1.
All the friends are divided into pairs. The pairings are given in a list pairs, where pairs[i] = [xi, yi] denotes xi is paired with yi and yi is paired with xi.
However, this pairing may cause some of the friends to be unhappy. A friend x is unhappy if x is paired with y and there exists a friend u who is paired with v but:
x prefers u over y, andu prefers x over v.Return the number of unhappy friends.
Example 1:
Input: n = 4, preferences = [[1, 2, 3], [3, 2, 0], [3, 1, 0], [1, 2, 0]], pairs = [[0, 1], [2, 3]] Output: 2 Explanation: Friend 1 is unhappy because: - 1 is paired with 0 but prefers 3 over 0, and - 3 prefers 1 over 2. Friend 3 is unhappy because: - 3 is paired with 2 but prefers 1 over 2, and - 1 prefers 3 over 0. Friends 0 and 2 are happy.
Example 2:
Input: n = 2, preferences = [[1], [0]], pairs = [[1, 0]] Output: 0 Explanation: Both friends 0 and 1 are happy.
Example 3:
Input: n = 4, preferences = [[1, 3, 2], [2, 3, 0], [1, 3, 0], [0, 2, 1]], pairs = [[1, 3], [0, 2]] Output: 4
Constraints:
2 <= n <= 500n is even.preferences.length == npreferences[i].length == n - 10 <= preferences[i][j] <= n - 1preferences[i] does not contain i.preferences[i] are unique.pairs.length == n/2pairs[i].length == 2xi != yi0 <= xi, yi <= n - 1Problem summary: You are given a list of preferences for n friends, where n is always even. For each person i, preferences[i] contains a list of friends sorted in the order of preference. In other words, a friend earlier in the list is more preferred than a friend later in the list. Friends in each list are denoted by integers from 0 to n-1. All the friends are divided into pairs. The pairings are given in a list pairs, where pairs[i] = [xi, yi] denotes xi is paired with yi and yi is paired with xi. However, this pairing may cause some of the friends to be unhappy. A friend x is unhappy if x is paired with y and there exists a friend u who is paired with v but: x prefers u over y, and u prefers x over v. Return the number of unhappy friends.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
4 [[1,2,3],[3,2,0],[3,1,0],[1,2,0]] [[0,1],[2,3]]
2 [[1],[0]] [[1,0]]
4 [[1,3,2],[2,3,0],[1,3,0],[0,2,1]] [[1,3],[0,2]]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1583: Count Unhappy Friends
class Solution {
public int unhappyFriends(int n, int[][] preferences, int[][] pairs) {
int[][] d = new int[n][n];
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n - 1; ++j) {
d[i][preferences[i][j]] = j;
}
}
int[] p = new int[n];
for (var e : pairs) {
int x = e[0], y = e[1];
p[x] = y;
p[y] = x;
}
int ans = 0;
for (int x = 0; x < n; ++x) {
int y = p[x];
for (int i = 0; i < d[x][y]; ++i) {
int u = preferences[x][i];
int v = p[u];
if (d[u][x] < d[u][v]) {
++ans;
break;
}
}
}
return ans;
}
}
// Accepted solution for LeetCode #1583: Count Unhappy Friends
func unhappyFriends(n int, preferences [][]int, pairs [][]int) (ans int) {
d := make([][]int, n)
for i := range d {
d[i] = make([]int, n)
}
for i := 0; i < n; i++ {
for j := 0; j < n-1; j++ {
d[i][preferences[i][j]] = j
}
}
p := make([]int, n)
for _, e := range pairs {
x, y := e[0], e[1]
p[x] = y
p[y] = x
}
for x := 0; x < n; x++ {
y := p[x]
for i := 0; i < d[x][y]; i++ {
u := preferences[x][i]
v := p[u]
if d[u][x] < d[u][v] {
ans++
break
}
}
}
return
}
# Accepted solution for LeetCode #1583: Count Unhappy Friends
class Solution:
def unhappyFriends(
self, n: int, preferences: List[List[int]], pairs: List[List[int]]
) -> int:
d = [{x: j for j, x in enumerate(p)} for p in preferences]
p = {}
for x, y in pairs:
p[x] = y
p[y] = x
ans = 0
for x in range(n):
y = p[x]
for i in range(d[x][y]):
u = preferences[x][i]
v = p[u]
if d[u][x] < d[u][v]:
ans += 1
break
return ans
// Accepted solution for LeetCode #1583: Count Unhappy Friends
struct Solution;
use std::collections::HashSet;
impl Solution {
fn unhappy_friends(n: i32, preferences: Vec<Vec<i32>>, pairs: Vec<Vec<i32>>) -> i32 {
let n = n as usize;
let mut list = vec![HashSet::new(); n];
for pair in pairs {
let x = pair[0] as usize;
let y = pair[1] as usize;
for &u in &preferences[x] {
let u = u as usize;
if u != y {
list[x].insert(u);
} else {
break;
}
}
for &v in &preferences[y] {
let v = v as usize;
if v != x {
list[y].insert(v);
} else {
break;
}
}
}
let mut res = 0;
for i in 0..n {
for j in 0..n {
if i != j && list[i].contains(&j) && list[j].contains(&i) {
res += 1;
break;
}
}
}
res
}
}
#[test]
fn test() {
let n = 4;
let preferences = vec_vec_i32![[1, 2, 3], [3, 2, 0], [3, 1, 0], [1, 2, 0]];
let pairs = vec_vec_i32![[0, 1], [2, 3]];
let res = 2;
assert_eq!(Solution::unhappy_friends(n, preferences, pairs), res);
let n = 2;
let preferences = vec_vec_i32![[1], [0]];
let pairs = vec_vec_i32![[1, 0]];
let res = 0;
assert_eq!(Solution::unhappy_friends(n, preferences, pairs), res);
let n = 4;
let preferences = vec_vec_i32![[1, 3, 2], [2, 3, 0], [1, 3, 0], [0, 2, 1]];
let pairs = vec_vec_i32![[1, 3], [0, 2]];
let res = 4;
assert_eq!(Solution::unhappy_friends(n, preferences, pairs), res);
}
// Accepted solution for LeetCode #1583: Count Unhappy Friends
function unhappyFriends(n: number, preferences: number[][], pairs: number[][]): number {
const d: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n - 1; ++j) {
d[i][preferences[i][j]] = j;
}
}
const p: number[] = Array(n).fill(0);
for (const [x, y] of pairs) {
p[x] = y;
p[y] = x;
}
let ans = 0;
for (let x = 0; x < n; ++x) {
const y = p[x];
for (let i = 0; i < d[x][y]; ++i) {
const u = preferences[x][i];
const v = p[u];
if (d[u][x] < d[u][v]) {
++ans;
break;
}
}
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
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.