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Formatted question description: https://leetcode.ca/all/1394.html

# 1394. Find Lucky Integer in an Array

Easy

## Description

Given an array of integers arr, a lucky integer is an integer which has a frequency in the array equal to its value.

Return a lucky integer in the array. If there are multiple lucky integers return the largest of them. If there is no lucky integer return -1.

Example 1:

Input: arr = [2,2,3,4]

Output: 2

Explanation: The only lucky number in the array is 2 because frequency[2] == 2.

Example 2:

Input: arr = [1,2,2,3,3,3]

Output: 3

Explanation: 1, 2 and 3 are all lucky numbers, return the largest of them.

Example 3:

Input: arr = [2,2,2,3,3]

Output: -1

Explanation: There are no lucky numbers in the array.

Example 4:

Input: arr = [5]

Output: -1

Example 5:

Input: arr = [7,7,7,7,7,7,7]

Output: 7

Constraints:

• 1 <= arr.length <= 500
• 1 <= arr[i] <= 500

## Solution

Use a map to store each number’s frequency. Then loop over all entries in the map. If a number’s frequency equals the number, update the largest lucky number. If a lucky number exists, return the largest one. If no lucky number is met, return -1.

• class Solution {
public int findLucky(int[] arr) {
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for (int num : arr) {
int count = map.getOrDefault(num, 0) + 1;
map.put(num, count);
}
int luckyNumber = -1;
Set<Integer> keySet = map.keySet();
for (int num : keySet) {
int count = map.get(num);
if (count == num)
luckyNumber = Math.max(luckyNumber, num);
}
return luckyNumber;
}
}

• class Solution:
def findLucky(self, arr: List[int]) -> int:
counter = Counter(arr)
ans = -1
for num, n in counter.items():
if num == n and ans < num:
ans = num
return ans


• class Solution {
public:
int findLucky(vector<int>& arr) {
int n = 510;
vector<int> counter(n);
for (int e : arr) ++counter[e];
int ans = -1;
for (int i = 1; i < n; ++i) {
if (i == counter[i] && ans < i) ans = i;
}
return ans;
}
};

• func findLucky(arr []int) int {
n := 510
counter := make([]int, n)
for _, e := range arr {
counter[e]++
}
ans := -1
for i := 1; i < n; i++ {
if i == counter[i] && ans < i {
ans = i
}
}
return ans
}

• function findLucky(arr: number[]): number {
const cnt = new Array(510).fill(0);
for (const x of arr) {
++cnt[x];
}
let ans = -1;
for (let x = 1; x < cnt.length; ++x) {
if (cnt[x] === x) {
ans = x;
}
}
return ans;
}


• class Solution {
/**
* @param Integer[] $arr * @return Integer */ function findLucky($arr) {
$max = -1; for ($i = 0; $i < count($arr); $i++) {$hashtable[$arr[$i]] += 1;
}
$keys = array_keys($hashtable);
for ($j = 0;$j < count($keys);$j++) {
if ($hashtable[$keys[$j]] ==$keys[$j])$max = max($max,$keys[$j]); } return$max;
}
}