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1394. Find Lucky Integer in an Array

Description

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

Return the largest lucky integer in the array. 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.

 

Constraints:

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

Solutions

Solution 1: Counting

We can use a hash table or array $cnt$ to count the occurrences of each number in $arr$, then traverse $cnt$ to find the largest $x$ that satisfies $cnt[x] = x$. If there is no such $x$, return $-1$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of $arr$.

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class Solution:
    def findLucky(self, arr: List[int]) -> int:
        cnt = Counter(arr)
        ans = -1
        for x, v in cnt.items():
            if x == v and ans < x:
                ans = x
        return ans
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class Solution {
    public int findLucky(int[] arr) {
        int[] cnt = new int[510];
        for (int x : cnt) {
            ++cnt[x];
        }
        int ans = -1;
        for (int x = 1; x < cnt.length; ++x) {
            if (cnt[x] == x) {
                ans = x;
            }
        }
        return ans;
    }
}
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class Solution {
public:
    int findLucky(vector<int>& arr) {
        int cnt[510];
        memset(cnt, 0, sizeof(cnt));
        for (int x : arr) {
            ++cnt[x];
        }
        int ans = -1;
        for (int x = 1; x < 510; ++x) {
            if (cnt[x] == x) {
                ans = x;
            }
        }
        return ans;
    }
};
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func findLucky(arr []int) int {
    cnt := [510]int{}
    for _, x := range arr {
        cnt[x]++
    }
    ans := -1
    for x := 1; x < len(cnt); x++ {
        if cnt[x] == x {
            ans = x
        }
    }
    return ans
}
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function findLucky(arr: number[]): number {
    const cnt = 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;
}
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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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