Find Lucky Integer in an Array

The task is to find a “lucky” integer from an array. An integer is considered lucky if its value is equal to its frequency in the array. The function should return the largest lucky integer or -1 if no such integer exists.

The approach to solve this problem is fairly straightforward:

  1. We first count the frequency of each number in the array. We can do this by iterating over the array and updating a dictionary where the keys are the numbers and the values are their frequencies.

  2. Then we check if each number is lucky i.e., if the number is equal to its frequency.

  3. While doing so, we keep track of the maximum lucky number.

Here is the Python code:

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class Solution:
    def findLucky(self, arr: List[int]) -> int:
        counter = collections.Counter(arr)
        max_lucky = -1
        for num, freq in counter.items():
            if num == freq:
                max_lucky = max(max_lucky, num)
        return max_lucky

This code starts by creating a counter dictionary from the list of integers arr. This dictionary (counter) maps each integer to its frequency in the list. It then iterates over each item in the dictionary. If the number (key) equals its frequency (value), it is considered a lucky number. The maximum lucky number (max_lucky) is updated if this lucky number is larger than the current maximum. If no lucky number is found, max_lucky will remain -1, which is the required output in such a case.

Brute Force

  1. Iterate over each distinct element in the array.
  2. For each distinct element, count its occurrence in the array.
  3. If an element’s value matches its count, compare it with the current largest lucky integer. If it’s larger, update the largest lucky integer.
  4. After all elements are processed, if no lucky integer is found, return -1. Otherwise, return the largest lucky integer.

Here’s how this can be implemented in Python:

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class Solution:
    def findLucky(self, arr: List[int]) -> int:
        max_lucky = -1
        for num in set(arr):
            if num == arr.count(num):
                max_lucky = max(max_lucky, num)
        return max_lucky

This solution iterates through each distinct element in the array and counts its occurrences using the arr.count(num) command. This command itself iterates over the entire array. As a result, this solution has a time complexity of O(n^2) for an array of size n, making it less efficient than the other solutions provided earlier.

Keep in mind that this solution might not be acceptable for large inputs due to its inefficiency. It’s usually better to use a more efficient algorithm like the ones previously mentioned. The brute force solution can be a good starting point for understanding the problem and coming up with more efficient solutions.

Q&A

First of all i would say do not go straight for the solution here but look out for the algorithms,i would like to give you some help here.

Use another array for the problem of lenght 501(counter) since the constarints are array length is not more than 500 and keep having a track of all elements and their counts in this array and then take another loop and then check if the counter array here contains same value as iterator variable and if yes take it one of the variable(since there is a possibility that we might find greater value than this and in the end return that variable.

Use HashMap for this,first go on adding all the elements and their values in it and then in the end if you find that the key and value are same store it’s value in the variable,return this variable in the end.Here you can use any other DS that you are more used to.

Sort the array and iterate through it from back and return the first element you find has value same as the number of times it has occured and return the same.

You can also use Heap here.

Use Bitwise and store the values in upper part of integer,then return the max one,this is most effective since it has time complexity of O(n) and space of O(1).

Your insights regarding the different approaches are indeed useful. Let me explain them one by one, using Python as the language:

  1. Array approach:

    We create a frequency array of size 501 (since the maximum number and maximum length of the array is 500). Then we traverse this frequency array in reverse. The first number whose index and value are the same is the largest lucky integer.

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    class Solution:
    def findLucky(self, arr: List[int]) -> int:
        freq = [0] * 501
        for num in arr:
            freq[num] += 1
        for i in range(500, -1, -1):
            if freq[i] == i:
                return i
        return -1

  2. HashMap approach:

    Here, we use a HashMap (in Python, a dict) to store the frequency of each number. We then find the maximum key which is equal to its value.

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    class Solution:
    def findLucky(self, arr: List[int]) -> int:
        counter = collections.Counter(arr)
        max_lucky = -1
        for num, freq in counter.items():
            if num == freq:
                max_lucky = max(max_lucky, num)
        return max_lucky

  3. Sorting approach:

    In this method, we first sort the array and then iterate through it from the end. We keep a count of the same elements and when the count is equal to the element, we return it.

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    class Solution:
    def findLucky(self, arr: List[int]) -> int:
        arr.sort(reverse=True)
        i = 0
        while i < len(arr):
            count = arr.count(arr[i])
            if count == arr[i]:
                return arr[i]
            i += count
        return -1

  4. Heap approach:

    In this method, we first create a frequency map. Then we push all items of the frequency map into a max-heap (priority queue where the key is the frequency). We pop the heap until we find a lucky integer or the heap is empty.

    Note: Python’s heapq module provides min-heap functionality. To make it behave like a max-heap, we multiply each key by -1.

  5. Bitwise approach:

    This approach is quite intricate and requires a deep understanding of bitwise operations. It’s also language-dependent as not all languages handle integers and bitwise operations the same way. It’s not commonly used and is beyond the scope of this response.

I hope these explanations provide you with an understanding of the various approaches you can take to solve this problem. It’s always beneficial to be aware of multiple strategies, as the most efficient one can depend on the exact constraints and requirements of your specific situation.

Linked List

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int findLucky(vector<int>& arr) {
    for (auto i = 0; i < arr.size(); ++i) {
        auto p = i, val = arr[i];
        while (val > 0 && val <= arr.size()) {
            auto n_val = arr[val - 1];
            arr[val - 1] = min(0, arr[val - 1]) - 1;
            if (val - 1 <= i || val - 1 == p)
                break;
            p = val - 1;
            val = n_val;
        }      
    }
    for (auto i = arr.size(); i > 0; --i)
        if (-arr[i - 1] == i)
            return i;
    return -1;
}

This function findLucky is designed to find a “lucky” integer in the given array. A “lucky” integer is an integer in the array that is equal to the frequency of its occurrence in the array. If there is more than one lucky integer, the function returns the largest one. If there is no lucky integer, the function returns -1.

The function uses a clever counting technique to find the frequency of occurrence of each integer in the array. Instead of using an additional data structure like a hash map to store the counts, it modifies the array in place.

Here’s a breakdown of the code:

  1. The first loop iterates over each element in the array. For each element, it performs a nested loop that redistributes the current element’s value to the position in the array corresponding to its value (minus 1, to account for zero-based indexing). It does this by decrementing the value at the target position.

    If the target position is less than or equal to the current position in the outer loop, or if it equals the previous target position, the nested loop breaks. This prevents an infinite loop in case of repeated values.

  2. The second loop iterates over the array from the end to the beginning. It checks each element to see if its absolute value (made negative in the first loop) equals its position in the array (plus 1, to account for zero-based indexing). If it does, that means the integer equal to the position is a lucky integer, and it returns that integer.

  3. If the second loop completes without finding a lucky integer, the function returns -1.

The time complexity of this function is linear, O(n), since it only makes a few passes over the array. The space complexity is constant, O(1), since it does not use any additional data structures.

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int findLucky(vector<int>& arr) {
    for (auto n : arr) {
        n &= 0xFFFF;
        if (n <= arr.size())
            arr[n - 1] += 0x10000;
    }
    for (auto i = arr.size(); i > 0; --i)
        if ((arr[i - 1] >> 16) == i)
            return i;
    return -1;
}

The function findLucky is trying to find a “lucky” integer in an array. A “lucky” integer is an integer whose value is equal to the frequency of its appearance in the array. If there are multiple “lucky” integers, the function returns the largest one. If there is no “lucky” integer, it returns -1.

The function uses bitwise operations to count the frequency of each integer in the array in place, without using any additional data structure. Here’s a detailed breakdown of the code:

  1. The first loop iterates over each element in the array. For each element:

    • It applies a bitwise AND operation with 0xFFFF to the element. This operation effectively keeps only the least significant 16 bits of the integer and sets the rest of the bits to 0. This is needed because later on, the function will use the higher 16 bits to store the frequency count of the element.

    • If the resulting number is less than or equal to the size of the array (which means it’s a valid index), it increments the 17th to 32nd bits of the array element at the index equal to the number minus 1. This is done by adding 0x10000 to the element, which is equivalent to incrementing the count by 1.

  2. The second loop iterates over the array from the end to the start. For each element:

    • It checks if the most significant 16 bits of the element (obtained by right-shifting the element by 16 bits) are equal to the index (plus 1, to account for zero-based indexing). If they are equal, it means the integer equal to the index is a “lucky” integer, and it returns that integer.

  3. If the second loop completes without finding a “lucky” integer, the function returns -1.

The time complexity of this function is linear, O(n), since it makes two passes over the array. The space complexity is constant, O(1), as it doesn’t use any additional data structures. This function assumes that the integers in the array are non-negative and fit within 16 bits.