Every Sublist Containing Unique Element - Google Top Interview Questions


Problem Statement :


Given a list of integers nums, return whether every sublist has at least 1 element in it which occurs exactly once in the sublist.

Note: the intended solution should run \mathcal {O}(n)O(n) for random inputs on average.

Constraints

n ≤ 100,000 where n is the length of nums
Example 1
Input
nums = [0, 2, 4, 2, 0]
Output
True
Explanation
Every sublist in nums has at least one element whose frequency is one. 
For example, the sublist [2, 4, 2] has 4. [0, 2, 4, 2] has 0 and 4.

Example 2

Input

nums = [2, 2, 0]

Output

False

Explanation

In the sublist [2, 2], there's not a unique element.


Solution :



title-img 




                        Solution in C++ :

void discretize(vector<int>& v) {
    vector<int> w = v;
    sort(w.begin(), w.end());
    w.resize(unique(w.begin(), w.end()) - w.begin());
    for (int& val : v) {
        int idx = lower_bound(w.begin(), w.end(), val) - w.begin();
        val = idx;
    }
}

int nxtidx[1000005];
int prvidx[1000005];
int lastloc[1000005];

bool possible(vector<int>& v, int lhs, int rhs) {
    while (true) {
        if (lhs >= rhs) return true;
        int a = lhs;
        int b = rhs;
        bool found = false;
        while (!found && a <= b) {
            if (nxtidx[a] > rhs && prvidx[a] < lhs) {
                if (!possible(v, lhs, a - 1)) return false;
                lhs = a + 1;
                found = true;
                break;
            }
            a++;
            if (prvidx[b] < lhs && nxtidx[b] > rhs) {
                if (!possible(v, b + 1, rhs)) return false;
                rhs = b - 1;
                found = true;
                break;
            }
            b--;
        }
        if (!found) return false;
    }
}

bool solve(vector<int>& nums) {
    if (nums.size() == 0) return true;
    discretize(nums);
    int n = nums.size();
    for (int i = 0; i < n; i++) nxtidx[i] = n;
    for (int i = 0; i < n; i++) prvidx[i] = -1;
    for (int i = 0; i < n; i++) lastloc[i] = -1;
    for (int i = 0; i < n; i++) {
        if (lastloc[nums[i]] >= 0) {
            prvidx[i] = lastloc[nums[i]];
            nxtidx[lastloc[nums[i]]] = i;
        }
        lastloc[nums[i]] = i;
    }
    return possible(nums, 0, nums.size() - 1);
}
                    




                        Solution in Python : 
                            
from collections import Counter, defaultdict


def sub_once(nums):
    if not nums:
        return True

    positions = defaultdict(list)
    for index, num in enumerate(nums):
        positions[num].append(index)

    pivots = [indexes[0] for indexes in positions.values() if len(indexes) == 1]

    if not pivots:
        return False

    if len(pivots) == len(nums):
        return True

    pivots.extend([-1, len(nums)])
    pivots.sort()

    return all(
        sub_once(nums[pivots[preindex] + 1 : pivot]) for preindex, pivot in enumerate(pivots[1:])
    )


class Solution:
    def solve(self, nums):
        return sub_once(nums)


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