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 :



title-img




                        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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