# 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)
sum = (sum + cur * (cur + 1) / 2) % MOD;
cur = 0;
}
}
if (cur > 0)
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```
```

## Subsequence Weighting

A subsequence of a sequence is a sequence which is obtained by deleting zero or more elements from the sequence. You are given a sequence A in which every element is a pair of integers i.e A = [(a1, w1), (a2, w2),..., (aN, wN)]. For a subseqence B = [(b1, v1), (b2, v2), ...., (bM, vM)] of the given sequence : We call it increasing if for every i (1 <= i < M ) , bi < bi+1. Weight(B) =

Meera teaches a class of n students, and every day in her classroom is an adventure. Today is drawing day! The students are sitting around a round table, and they are numbered from 1 to n in the clockwise direction. This means that the students are numbered 1, 2, 3, . . . , n-1, n, and students 1 and n are sitting next to each other. After letting the students draw for a certain period of ti

## Mr. X and His Shots

A cricket match is going to be held. The field is represented by a 1D plane. A cricketer, Mr. X has N favorite shots. Each shot has a particular range. The range of the ith shot is from Ai to Bi. That means his favorite shot can be anywhere in this range. Each player on the opposite team can field only in a particular range. Player i can field from Ci to Di. You are given the N favorite shots of M

## Jim and the Skyscrapers

Jim has invented a new flying object called HZ42. HZ42 is like a broom and can only fly horizontally, independent of the environment. One day, Jim started his flight from Dubai's highest skyscraper, traveled some distance and landed on another skyscraper of same height! So much fun! But unfortunately, new skyscrapers have been built recently. Let us describe the problem in one dimensional space

## Palindromic Subsets

Consider a lowercase English alphabetic letter character denoted by c. A shift operation on some c turns it into the next letter in the alphabet. For example, and ,shift(a) = b , shift(e) = f, shift(z) = a . Given a zero-indexed string, s, of n lowercase letters, perform q queries on s where each query takes one of the following two forms: 1 i j t: All letters in the inclusive range from i t

## Counting On a Tree

Taylor loves trees, and this new challenge has him stumped! Consider a tree, t, consisting of n nodes. Each node is numbered from 1 to n, and each node i has an integer, ci, attached to it. A query on tree t takes the form w x y z. To process a query, you must print the count of ordered pairs of integers ( i , j ) such that the following four conditions are all satisfied: the path from n