Given two arrays **S[]** and **E[]** of size** N** denoting starting and closing time of the shops and an integer value** K** denoting the number of people, the task is to find out the maximum number of shops they can visit in total if they visit each shop optimally based on the following conditions:

- A shop can be visited by only one person
- A person cannot visit another shop if its timing collide with it

**Examples:**

Input:S[] = {1, 8, 3, 2, 6}, E[] = {5, 10, 6, 5, 9}, K = 2Output:4Explanation:One possible solution can be that first person visits the 1st and 5th shops meanwhile second person will visit 4th and 2nd shops.

Input:S[] = {1, 2, 3}, E[] = {3, 4, 5}, K = 2Output:3Explanation:One possible solution can be that first person visits the 1st and 3rd shops meanwhile second person will visit 2nd shop.

**Approach: **This problem can be solved using the greedy technique called activity selection and sorting. In the activity selection problem, only one person performs the activity but here **K** persons are available for one activity. To manage the availability of one person a multiset is used.

Follow the steps below to solve the problem:

- Initialize an array
**a[]**of pairs and store the pair**{S[i], E[i]}**for each index**i**. - Sort the array
**a[]**according to the ending time. - Initialize a multiset
**st**to store the persons with ending time of shop they are currently visiting. - Initialize a variable
**ans**with**0**to store the final result. - Traverse each pair of array
**a[]**,- If a person is available i.e
**a[i].first**is greater than or equal to the ending time of any person in the multiset**st**, then increment the**count**by**1**and update the ending time of that element with new**a[i].second**. - Otherwise, continue checking the next pairs.

- If a person is available i.e
- Finally, print the result as
**count**.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Comparator ` `bool` `compareBy(` `const` `pair<` `int` `, ` `int` `>& a, ` ` ` `const` `pair<` `int` `, ` `int` `>& b) ` `{ ` ` ` `if` `(a.second != b.second) ` ` ` `return` `a.second < b.second; ` ` ` `return` `a.first < b.first; ` `} ` `// Function to find maximum shops ` `// that can be visited by K persons ` `int` `maximumShops(` `int` `* opening, ` `int` `* closing, ` ` ` `int` `n, ` `int` `k) ` `{ ` ` ` `// Store opening and closing ` ` ` `// time of shops ` ` ` `pair<` `int` `, ` `int` `> a[n]; ` ` ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` `a[i].first = opening[i]; ` ` ` `a[i].second = closing[i]; ` ` ` `} ` ` ` ` ` `// Sort the pair of array ` ` ` `sort(a, a + n, compareBy); ` ` ` ` ` `// Stores the result ` ` ` `int` `count = 0; ` ` ` ` ` `// Stores current number of persons visting ` ` ` `// some shop with their ending time ` ` ` `multiset<` `int` `> st; ` ` ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` ` ` `// Check if current shop can be ` ` ` `// assigned to a person who's ` ` ` `// already visiting any other shop ` ` ` `bool` `flag = ` `false` `; ` ` ` ` ` `if` `(!st.empty()) { ` ` ` ` ` `auto` `it = st.upper_bound(a[i].first); ` ` ` ` ` `if` `(it != st.begin()) { ` ` ` `it--; ` ` ` ` ` `// Checks if there is any person whose ` ` ` `// closing time <= current shop opening ` ` ` `// time ` ` ` `if` `(*it <= a[i].first) { ` ` ` ` ` `// Erase previous shop visited by the ` ` ` `// person satisfying the condition ` ` ` `st.erase(it); ` ` ` ` ` `// Insert new closing time of current ` ` ` `// shop for the person satisfying ṭhe ` ` ` `// condition ` ` ` `st.insert(a[i].second); ` ` ` ` ` `// Increment the count by one ` ` ` `count++; ` ` ` ` ` `flag = ` `true` `; ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` ` ` `// In case if no person have closing ` ` ` `// time <= current shop opening time ` ` ` `// but there are some persons left ` ` ` `if` `(st.size() < k && flag == ` `false` `) { ` ` ` `st.insert(a[i].second); ` ` ` `count++; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Finally print the ans ` ` ` `return` `count; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` ` ` `// Given starting and ending time ` ` ` `int` `S[] = { 1, 8, 3, 2, 6 }; ` ` ` `int` `E[] = { 5, 10, 6, 5, 9 }; ` ` ` ` ` `// Given K and N ` ` ` `int` `K = 2, N = ` `sizeof` `(S) ` ` ` `/ ` `sizeof` `(S[0]); ` ` ` ` ` `// Function call ` ` ` `cout << maximumShops(S, E, N, K) << endl; ` `}` |

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## Python3

`# Python3 program for the above approach ` `from` `bisect ` `import` `bisect_left ` ` ` `# Function to find maximum shops ` `# that can be visited by K persons ` `def` `maximumShops(opening, closing, n, k): ` ` ` ` ` `# Store opening and closing ` ` ` `# time of shops ` ` ` `a ` `=` `[[` `0` `, ` `0` `] ` `for` `i ` `in` `range` `(n)] ` ` ` ` ` `for` `i ` `in` `range` `(n): ` ` ` `a[i][` `0` `] ` `=` `opening[i] ` ` ` `a[i][` `1` `] ` `=` `closing[i] ` ` ` ` ` `# Sort the pair of array ` ` ` `a ` `=` `sorted` `(a) ` ` ` ` ` `# Stores the result ` ` ` `count ` `=` `1` ` ` ` ` `# Stores current number of persons visting ` ` ` `# some shop with their ending time ` ` ` `st ` `=` `{} ` ` ` `for` `i ` `in` `range` `(n): ` ` ` ` ` `# Check if current shop can be ` ` ` `# assigned to a person who's ` ` ` `# already visiting any other shop ` ` ` `flag ` `=` `False` ` ` ` ` `if` `(` `len` `(st) ` `=` `=` `0` `): ` ` ` `ar ` `=` `list` `(st.keys()) ` ` ` ` ` `it ` `=` `bisect_left(ar, a[i][` `0` `]) ` ` ` ` ` `if` `(it !` `=` `0` `): ` ` ` `it ` `-` `=` `1` ` ` ` ` `# Checks if there is any person whose ` ` ` `# closing time <= current shop opening ` ` ` `# time ` ` ` `if` `(ar[it] <` `=` `a[i][` `0` `]): ` ` ` ` ` `# Erase previous shop visited by the ` ` ` `# person satisfying the condition ` ` ` `del` `st[it] ` ` ` ` ` `# Insert new closing time of current ` ` ` `# shop for the person satisfying ṭhe ` ` ` `# condition ` ` ` `st[a[i][` `1` `]] ` `=` `1` ` ` ` ` `# Increment the count by one ` ` ` `count ` `+` `=` `1` ` ` `flag ` `=` `True` ` ` ` ` `# In case if no person have closing ` ` ` `# time <= current shop opening time ` ` ` `# but there are some persons left ` ` ` `if` `(` `len` `(st) < k ` `and` `flag ` `=` `=` `False` `): ` ` ` `st[a[i][` `1` `]] ` `=` `1` ` ` `count ` `+` `=` `1` ` ` ` ` `# Finally pr the ans ` ` ` `return` `count ` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` ` ` `# Given starting and ending time ` ` ` `S ` `=` `[` `1` `, ` `8` `, ` `3` `, ` `2` `, ` `6` `] ` ` ` `E ` `=` `[` `5` `, ` `10` `, ` `6` `, ` `5` `, ` `9` `] ` ` ` ` ` `# Given K and N ` ` ` `K,N ` `=` `2` `, ` `len` `(S) ` ` ` ` ` `# Function call ` ` ` `print` `(maximumShops(S, E, N, K)) ` ` ` ` ` `# This code is contributed by mohit kumar 29` |

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**Output:**

4

**Time complexity:** O(NlogN)**Auxiliary Space: **O(N)

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