**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 :

` ````
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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