Sublist with Largest Min-Length Product - Google Top Interview Questions


Problem Statement :


You are given a list of integers nums and an integer pos. Find a sublist A of nums that includes the index (0-indexed) pos such that min(A) * A.length is maximized and return the value.

Constraints

n ≤ 100,000 where n is the length of nums

Example 1

Input

nums = [-1, 1, 4, 3]

pos = 3

Output

6

Explanation

The best sublist is [4, 3]. Since min(4, 3) = 3 and its length is 2 we have 3 * 2 = 6.


Solution :



title-img 




                        Solution in C++ :

int par[100005];
int sz[100005];
bool active[100005];
int find(int x) {
    return par[x] == x ? x : (par[x] = find(par[x]));
}
void merge(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) return;
    par[x] = y;
    sz[y] += sz[x];
}

int solve(vector<int>& nums, int pos) {
    if (nums[pos] < 0) return nums[pos];
    int n = nums.size();
    vector<pair<int, int>> v;
    for (int i = 0; i < n; i++) v.emplace_back(nums[i], i);
    sort(v.rbegin(), v.rend());
    int ret = -1e9;
    for (int i = 0; i < n; i++) {
        par[i] = i;
        sz[i] = 1;
        active[i] = false;
    }
    for (auto [value, index] : v) {
        active[index] = true;
        if (index > 0 && active[index - 1]) merge(index - 1, index);
        if (index + 1 < n && active[index + 1]) merge(index, index + 1);
        if (find(pos) == find(index)) ret = max(ret, sz[find(pos)] * value);
    }
    return ret;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    static int NEG = -100000000;
    public int solve(int[] nums, int pos) {
        int l = pos;
        int r = pos;
        int currMin = nums[pos];

        int ret = NEG;

        for (int i = 1; i <= nums.length; i++) {
            ret = Math.max(ret, currMin * i);

            int left = l - 1 >= 0 ? nums[l - 1] : NEG;
            int right = r + 1 < nums.length ? nums[r + 1] : NEG;

            if (left > right) {
                currMin = Math.min(currMin, left);
                l--;
            } else {
                currMin = Math.min(currMin, right);
                r++;
            }
        }
        return ret;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, A, pos):
        NINF = float("-inf")
        ans = m = A[pos]
        i = j = pos

        for _ in range(len(A) - 1):
            left = A[i - 1] if i - 1 >= 0 else NINF
            right = A[j + 1] if j + 1 < len(A) else NINF

            if left >= right:
                i -= 1
                m = min(m, A[i])
            else:
                j += 1
                m = min(m, A[j])

            ans = max(ans, m * (j - i + 1))

        return ans


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