List Consecutive Split - Google Top Interview Questions


Problem Statement :


Given a list of integers nums, and an integer k, return whether the list can be split into lists where each list contains k integers and is consecutively increasing.

Constraints

n ≤ 100,000 where n is the length of nums.

Example 1

Input

nums = [3, 2, 3, 4, 5, 1]

k = 3

Output

True

Explanation

We can split the list into [1, 2, 3] and [3, 4, 5]


Solution :



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                        Solution in C++ :

bool solve(vector<int>& nums, int k) {
    map<int, int> m;
    for (int n : nums) m[n]++;
    while (m.size()) {
        auto [x, y] = *m.begin();
        if (y == 0) {
            m.erase(x);
            continue;
        }
        for (int i = x; i < x + k; i++) {
            if (m[i] < y) {
                return false;
            }
            m[i] -= y;
        }
    }
    return m.size() == 0;
}
                    

                        Solution in Java :

import java.util.*;

class Solution {
    public boolean solve(int[] nums, int k) {
        if (nums == null || nums.length == 0 || k > nums.length)
            return false;
        Map<Integer, Integer> freqMap = new HashMap();
        for (int num : nums) freqMap.put(num, freqMap.getOrDefault(num, 0) + 1);
        PriorityQueue<Integer> pq = new PriorityQueue<>();
        for (int key : freqMap.keySet()) pq.offer(key);

        while (!pq.isEmpty() && pq.size() >= k) {
            int curr = -1;
            List<Integer> temp = new ArrayList();
            for (int i = 1; i <= k; i++) {
                int num = pq.poll();
                temp.add(num);
                if (freqMap.get(num) == 1)
                    freqMap.remove(num);
                else
                    freqMap.put(num, freqMap.get(num) - 1);
                if (curr == -1) {
                    curr = num;
                } else {
                    if (num != curr + 1)
                        return false;
                    curr = num;
                }
            }
            for (int i = 0; i < temp.size(); i++) {
                if (freqMap.containsKey(temp.get(i)))
                    pq.offer(temp.get(i));
            }
        }
        return pq.isEmpty();
    }
}
                    

                        Solution in Python : 
                            
class Solution:
    def solve(self, nums, k):
        fre = Counter(nums)
        # greedy solution, take the smallest then try
        vals = list(fre.keys())
        heapq.heapify(vals)

        # every iteration i make a new list
        while len(vals):
            # let us start with the minimum element
            start = vals[0]
            # now create a list starting with 'start'
            for i in range(k):
                if fre[start + i]:
                    fre[start + i] -= 1
                else:
                    return False
            # pop any elements which are not remaining
            while len(vals) and fre[vals[0]] == 0:
                heapq.heappop(vals)

        return True
                    

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