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

1394. Find Lucky Integer in an Array

Level

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;
        }
    }
    

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