# Count Unhappy Friends SOLUTION LEETCODE

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### Count Unhappy Friends SOLUTION

You are given top notch of inclinations for n companions, where n is in every case even.

For every individual I, preferences[i] contains a rundown of companions arranged in the request for inclination. At the end of the day, a companion prior in the rundown is more favored than a companion later in the rundown. Companions in each rundown are signified by whole numbers from 0 to n-1.

All the companions are partitioned into sets. The pairings are given in a rundown sets, where pairs[i] = [xi, yi] signifies xi is combined with yi and yi is matched with xi.

Nonetheless, this blending may make a portion of the companions be miserable. A companion x is troubled if x is combined with y and there exists a companion u who is matched with v however:

x lean towards u over y, and

u lean towards x over v.

Return the quantity of miserable companions.

Model 1:

Info: n = 4, inclinations = [[1, 2, 3], [3, 2, 0], [3, 1, 0], [1, 2, 0]], sets = [[0, 1], [2, 3]]

Yield: 2

Clarification:

Companion 1 is despondent in light of the fact that:

– 1 is combined with 0 yet favors 3 more than 0, and

– 3 favors 1 more than 2.

Companion 3 is despondent on the grounds that:

– 3 is combined with 2 yet favors 1 more than 2, and

– 1 favors 3 more than 0.

Companions 0 and 2 are upbeat.

Model 2:

Information: n = 2, inclinations = [[1], [0]], sets = [[1, 0]]

Yield: 0

Clarification: Both companions 0 and 1 are upbeat.

Model 3:

Information: n = 4, inclinations = [[1, 3, 2], [2, 3, 0], [1, 3, 0], [0, 2, 1]], sets = [[1, 3], [0, 2]]

Yield: 4

Imperatives:

2 <= n <= 500

n is even.

preferences.length == n

preferences[i].length == n – 1

0 <= preferences[i][j] <= n – 1

preferences[i] doesn’t contain I.

All qualities in preferences[i] are novel.

pairs.length == n/2

pairs[i].length == 2

xi != yi

0 <= xi, yi <= n – 1

Every individual is contained in precisely one sets.