**Non-Overlapping Pairs of Sublists - Google Top Interview Questions**

### Problem Statement :

Given a list of integers nums and a positive integer k, return the number of pairs of non-overlapping sublists such that all the elements in each sublist have value greater than or equal to k. Mod the result by 10 ** 9 + 7. Constraints n ≤ 100,000 where n is the length of nums 1 ≤ k Example 1 Input nums = [3, 4, 4, 9] k = 4 Output 5 Explanation These are the pairs of sublists: [4], [9] [4], [9] (using other 4) [4], [4] [4, 4], [9] [4], [4, 9]

### Solution :

` ````
Solution in C++ :
const int MOD = 1000000007;
long long self(int len) {
long long ret = 0;
for (int i = 0; i < len; i++) {
long long lhs = i + 1;
long long leftover = len - 1 - i;
long long rhs = leftover * (leftover + 1) / 2;
ret += lhs * rhs;
ret %= MOD;
}
return ret;
}
int solve(vector<int>& nums, int k) {
vector<int> lens;
long long ret = 0;
for (int i = 0; i < nums.size();) {
if (nums[i] < k) {
i++;
continue;
}
int j = i + 1;
while (j < nums.size() && nums[j] >= k) j++;
lens.push_back(j - i);
i = j;
}
for (int out : lens) {
ret += self(out);
}
long long inc = 0;
for (int len : lens) {
long long currentOption = (len * (len + 1LL)) / 2;
currentOption %= MOD;
long long cand = inc;
cand *= currentOption;
ret += cand;
ret %= MOD;
inc += currentOption;
inc %= MOD;
}
return ret % MOD;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] nums, int k) {
long MOD = 1000000007L;
ArrayList<Long> arr = new ArrayList<Long>();
long cur = 0;
long sum = 0;
for (int n : nums) {
if (n >= k) {
cur++;
} else {
if (cur > 0)
arr.add(cur);
sum = (sum + cur * (cur + 1) / 2) % MOD;
cur = 0;
}
}
if (cur > 0)
arr.add(cur);
sum = (sum + cur * (cur + 1) / 2) % MOD;
long ans = 0;
// the sublists come from different runs of the array
for (long n : arr) {
long d = (n * (n + 1) / 2) % MOD;
sum = (sum - d + MOD) % MOD;
ans = (ans + d * sum) % MOD;
}
// the sublists come from the same run of the array
for (long n : arr) {
if (n == 1)
continue;
// using stars and bars, the formula for size n is (n+2)C4.
long num = 1L;
for (int i = -1; i <= 2; i++) num = (num * (n + i)) % MOD;
long denInv = 41666667L; // mod inverse of 24
ans = (ans + num * denInv) % MOD;
}
return (int) ans;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums, k):
MOD = 10 ** 9 + 7
n = len(nums)
suffix = [0] * (n + 1)
end = n - 1
for start in range(n - 1, -1, -1):
num = nums[start]
if num < k:
end = start - 1
length = end - start + 1
suffix[start] += suffix[start + 1] + length
suffix[start] %= MOD
res = 0
start = 0
for end in range(n):
num = nums[end]
if num < k:
start = end + 1
length = end - start + 1
res += length * suffix[end + 1]
res %= MOD
return res
```

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