Every Sublist Min Sum- Google Top Interview Questions
Problem Statement :
You are given a list of integers nums. Return the sum of min(x) for every sublist x in nums. Mod the result by 10 ** 9 + 7. Constraints n ≤ 100,000 where n is the length of nums Example 1 Input nums = [1, 2, 4, 3] Output 20 Explanation We have the following sublists and their mins: min([1]) = 1 min([1, 2]) = 1 min([1, 2, 4]) = 1 min([1, 2, 4, 3]) = 1 min([2]) = 2 min([2, 4]) = 2 min([2, 4, 3]) = 2 min([4]) = 4 min([4, 3]) = 3 min([3]) = 3
Solution :
Solution in C++ :
int solve(vector<int>& nums) {
int n = nums.size();
int pref[n + 1] = {};
stack<int> s; // Initialising prefix array and stack
int ans = 0, j = 0, mod = 1e9 + 7;
for (int i = 0; i < n; i++) {
while (!s.empty() && nums[s.top()] >= nums[i])
s.pop(); // iterating back till we find an index where the minimum of the list if we go
// any further left will change
if (s.empty())
j = -1; // we are calculating j just for calculating the count for which nums[i] will
// appear as minimum in the sublist we have created by going left
else
j = s.top();
s.push(i);
pref[i + 1] = (pref[j + 1] + (i - j) * nums[i]) %
mod; // here maintaining prefix array--Read the implementation part
ans = (ans + pref[i + 1]) % mod;
}
return ans;
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] nums) {
int n = nums.length;
int[] next_smaller = new int[n];
int[] prev_smaller = new int[n];
Arrays.fill(next_smaller, n);
Arrays.fill(prev_smaller, -1);
// compute and fill in the next_smaller
Stack<Integer> stack = new Stack();
for (int i = 0; i < n; i++) {
int cur = nums[i];
while (stack.size() > 0 && cur < nums[stack.peek()]) {
next_smaller[stack.pop()] = i;
}
stack.push(i);
}
// compute and fill in the prev_smaller
stack = new Stack();
for (int i = 0; i < n; i++) {
int cur = nums[i];
while (stack.size() > 0 && nums[stack.peek()] > cur) {
stack.pop();
}
if (stack.size() > 0)
prev_smaller[i] = stack.peek();
stack.push(i);
}
// compute the answer
long res = 0;
long mod = (long) 1e9 + 7;
for (int i = 0; i < n; i++) {
int subarray_min = nums[i];
int left_count = i - prev_smaller[i]; // number of greater elements on the left
int right_count =
next_smaller[i] - i; // number of greater than or equal to elements on right
res = res + (subarray_min * (left_count * right_count)) % mod;
res = res % mod;
}
return (int) res;
}
}
Solution in Python :
class Solution:
def solve(self, nums):
ans = 0
s = [] # stack of (index, value, sum)
currentSum = 0
for index, value in enumerate(nums):
while s and value <= s[-1][1]:
currentSum -= s[-1][2]
s.pop()
if not s:
s.append([index, value, (index + 1) * value])
else:
s.append([index, value, (index - s[-1][0]) * value])
currentSum += s[-1][2]
ans += currentSum
return ans % (10 ** 9 + 7)
View More Similar Problems
Kundu and Tree
Kundu is true tree lover. Tree is a connected graph having N vertices and N-1 edges. Today when he got a tree, he colored each edge with one of either red(r) or black(b) color. He is interested in knowing how many triplets(a,b,c) of vertices are there , such that, there is atleast one edge having red color on all the three paths i.e. from vertex a to b, vertex b to c and vertex c to a . Note that
View Solution →Super Maximum Cost Queries
Victoria has a tree, T , consisting of N nodes numbered from 1 to N. Each edge from node Ui to Vi in tree T has an integer weight, Wi. Let's define the cost, C, of a path from some node X to some other node Y as the maximum weight ( W ) for any edge in the unique path from node X to Y node . Victoria wants your help processing Q queries on tree T, where each query contains 2 integers, L and
View Solution →Contacts
We're going to make our own Contacts application! The application must perform two types of operations: 1 . add name, where name is a string denoting a contact name. This must store name as a new contact in the application. find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting partial with and print the co
View Solution →No Prefix Set
There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio
View Solution →Cube Summation
You are given a 3-D Matrix in which each block contains 0 initially. The first block is defined by the coordinate (1,1,1) and the last block is defined by the coordinate (N,N,N). There are two types of queries. UPDATE x y z W updates the value of block (x,y,z) to W. QUERY x1 y1 z1 x2 y2 z2 calculates the sum of the value of blocks whose x coordinate is between x1 and x2 (inclusive), y coor
View Solution →Direct Connections
Enter-View ( EV ) is a linear, street-like country. By linear, we mean all the cities of the country are placed on a single straight line - the x -axis. Thus every city's position can be defined by a single coordinate, xi, the distance from the left borderline of the country. You can treat all cities as single points. Unfortunately, the dictator of telecommunication of EV (Mr. S. Treat Jr.) do
View Solution →