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3346. Maximum Frequency of an Element After Performing Operations I

Description

You are given an integer array nums and two integers k and numOperations.

You must perform an operation numOperations times on nums, where in each operation you:

  • Select an index i that was not selected in any previous operations.
  • Add an integer in the range [-k, k] to nums[i].

Return the maximum possible frequency of any element in nums after performing the operations.

 

Example 1:

Input: nums = [1,4,5], k = 1, numOperations = 2

Output: 2

Explanation:

We can achieve a maximum frequency of two by:

  • Adding 0 to nums[1]. nums becomes [1, 4, 5].
  • Adding -1 to nums[2]. nums becomes [1, 4, 4].

Example 2:

Input: nums = [5,11,20,20], k = 5, numOperations = 1

Output: 2

Explanation:

We can achieve a maximum frequency of two by:

  • Adding 0 to nums[1].

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 105
  • 0 <= k <= 105
  • 0 <= numOperations <= nums.length

Solutions

Solution 1

  • class Solution {
    public int maxFrequency(int[] nums, int k, int numOperations) {
        Map<Integer, Integer> cnt = new HashMap<>();
        TreeMap<Integer, Integer> d = new TreeMap<>();
        for (int x : nums) {
            cnt.merge(x, 1, Integer::sum);
            d.putIfAbsent(x, 0);
            d.merge(x - k, 1, Integer::sum);
            d.merge(x + k + 1, -1, Integer::sum);
        }
        int ans = 0, s = 0;
        for (var e : d.entrySet()) {
            int x = e.getKey(), t = e.getValue();
            s += t;
            ans = Math.max(ans, Math.min(s, cnt.getOrDefault(x, 0) + numOperations));
        }
        return ans;
    }
}


  • class Solution {
public:
    int maxFrequency(vector<int>& nums, int k, int numOperations) {
        unordered_map<int, int> cnt;
        map<int, int> d;

        for (int x : nums) {
            cnt[x]++;
            d[x];
            d[x - k]++;
            d[x + k + 1]--;
        }

        int ans = 0, s = 0;
        for (const auto& [x, t] : d) {
            s += t;
            ans = max(ans, min(s, cnt[x] + numOperations));
        }

        return ans;
    }
};


  • class Solution:
    def maxFrequency(self, nums: List[int], k: int, numOperations: int) -> int:
        cnt = defaultdict(int)
        d = defaultdict(int)
        for x in nums:
            cnt[x] += 1
            d[x] += 0
            d[x - k] += 1
            d[x + k + 1] -= 1
        ans = s = 0
        for x, t in sorted(d.items()):
            s += t
            ans = max(ans, min(s, cnt[x] + numOperations))
        return ans


  • func maxFrequency(nums []int, k int, numOperations int) (ans int) {
	cnt := make(map[int]int)
	d := make(map[int]int)
	for _, x := range nums {
		cnt[x]++
		d[x] = d[x]
		d[x-k]++
		d[x+k+1]--
	}

	s := 0
	keys := make([]int, 0, len(d))
	for key := range d {
		keys = append(keys, key)
	}
	sort.Ints(keys)
	for _, x := range keys {
		s += d[x]
		ans = max(ans, min(s, cnt[x]+numOperations))
	}

	return
}


  • function maxFrequency(nums: number[], k: number, numOperations: number): number {
    const cnt: Record<number, number> = {};
    const d: Record<number, number> = {};
    for (const x of nums) {
        cnt[x] = (cnt[x] || 0) + 1;
        d[x] = d[x] || 0;
        d[x - k] = (d[x - k] || 0) + 1;
        d[x + k + 1] = (d[x + k + 1] || 0) - 1;
    }
    let [ans, s] = [0, 0];
    const keys = Object.keys(d)
        .map(Number)
        .sort((a, b) => a - b);
    for (const x of keys) {
        s += d[x];
        ans = Math.max(ans, Math.min(s, (cnt[x] || 0) + numOperations));
    }

    return ans;
}

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