Boyer–Moore majority vote algorithm

_posts/2024-05-13-Boyer–Moore-majority-vote-algorithm.md

@@ -0,0 +1,75 @@
+---
+layout: post
+title: Boyer–Moore majority vote algorithm
+date: 2024-05-13 13:30:00 +0800
+categories: algorithm
+tags: vote
+published: true
+---
+
+* content
+{:toc}
+
+[Boyer–Moore majority vote algorithm](https://www.cs.utexas.edu/users/moore/best-ideas/mjrty/index.html){:target="_blank"}
+
+## problem
+
+[Kotlin Heroes: Practice 10 - Problem A](https://codeforces.com/contest/1959/problem/A){:target="_blank"}
+
+You are given an array 𝑎 consisting of 𝑛(𝑛≥3) positive integers. 
+It is known that in this array, all the numbers except one are the same 
+(for example, in the array [4,11,4,4] all numbers except one are equal to 4).
+
+Print the index of the element that does not equal others. The numbers in the array are numbered from one.
+
+**Input**<br>
+The first line contains a single integer 𝑡(1≤𝑡≤100). Then 𝑡 test cases follow.<br>
+The first line of each test case contains a single integer 𝑛(3≤𝑛≤100) — the length of the array 𝑎.<br>
+The second line of each test case contains 𝑛 integers 𝑎1,𝑎2,…,𝑎𝑛(1≤𝑎𝑖≤100).<br>
+It is guaranteed that all the numbers except one in the 𝑎 array are the same.
+
+**Output**<br>
+For each test case, output a single integer — the index of the element that is not equal to others.
+
+**Example**
+```
+input
+4
+4
+11 13 11 11
+5
+1 4 4 4 4
+10
+3 3 3 3 10 3 3 3 3 3
+3
+20 20 10
+
+output
+2
+1
+5
+3
+```
+
+## solution
+
+```kotlin
+fun main() {
+    repeat(readln().toInt()) { println(solve()) }
+}
+
+fun solve(): Int {
+    readln()
+    val array = readln().split(' ').map { it.toInt() }.toIntArray()
+    val list = mutableListOf<Int>()
+    repeat(3) {
+        if (list.isEmpty() || list[0] == array[it]) {
+            list.add(array[it])
+        } else {
+            list.removeAt(0)
+        }
+    }
+    val majority = list[0]
+    return array.indexOfFirst { it != majority } + 1
+}
+```