Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Find the Kth Smallest Sum of a Matrix With Sorted Rows in C++
Suppose we have one m * n matrix called mat, and an integer k, mat has its rows sorted in nondecreasing order. We can choose exactly one element from each row to form an array. We have to find the Kth smallest array sum among all possible arrays.
So, if the input is like mat = [[1,3,11],[2,4,6]]
| 1 | 3 | 1 1 |
| 2 | 4 | 6 |
and k = 5, then the output will be 7, as when we choose one element from each row the first k smallest sums are [1,2], [1,4], [3,2], [3,4], [1,6]. here the 5th sum is 7.
To solve this, we will follow these steps −
Define one priority queue pq
Define one 2D array m
Define a function update(), this will take an array v, i, ok initialize it with false,
if i is same as size of v, then −
if ok is false, then −
return
return
for initialize j := 0, when j < size of v, update (increase j by 1), do −
sum := sum + m[j, v[j]]
Define an array temp and copy v into it
insert sum into temp at the beginning
insert temp into pq
return
(increase v[i] by 1)
if ok is false and v[i] < z, then −
update(v, i + 1, true)
update(v, i + 1, true)
update(v, i + 1, ok)
From the main method do the following −
m :+ given matrix
ret := 0
n := row count of m
z := column count of m
for initialize i := 0, when i < n, update (increase i by 1), do −
ret := ret + m[i, 0]
Define an array temp of size n
insert ret into temp at the beginning
insert temp into pq
Define one set s
while k is non-zero, decrease k by 1 in each iteration, do −
Define an array temp = top of pq
delete element from pq
insert temp into s
ret := temp[0]
delete first element of temp from temp
update(temp, 0)
while (not pq is empty and top element of pq is member of s), do −
delete element from pq
return ret
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
struct Cmp{
bool operator()(vector <int>& a, vector <int>& b) {
return !(a[0] < b[0]);
}
};
class Solution {
public:
priority_queue<vector<int>, vector<vector<int> >, Cmp> pq;
vector<vector<int> > m;
int z;
void update(vector<int>& v, int i, bool ok = false){
if (i == v.size()) {
if (!ok)
return;
int sum = 0;
for (int j = 0; j < v.size(); j++) {
sum += m[j][v[j]];
}
vector<int> temp(v.begin(), v.end());
temp.insert(temp.begin(), sum);
pq.push(temp);
return;
}
v[i]++;
if (!ok && v[i] < z)
update(v, i + 1, true);
v[i]--;
update(v, i + 1, ok);
}
int kthSmallest(vector<vector<int> >& m, int k){
this->m = m;
int ret = 0;
int n = m.size();
z = m[0].size();
for (int i = 0; i < n; i++) {
ret += m[i][0];
}
vector<int> temp(n);
temp.insert(temp.begin(), ret);
pq.push(temp);
set<vector<int> > s;
while (k--) {
vector<int> temp = pq.top();
pq.pop();
s.insert(temp);
ret = temp[0];
temp.erase(temp.begin());
update(temp, 0);
while (!pq.empty() && s.count(pq.top())) {
pq.pop();
}
}
return ret;
}
};
main(){
Solution ob;
vector<vector<int>> v = {{1,3,11},{2,4,6}};
cout << (ob.kthSmallest(v, 5));
}Input
{{1,3,11},{2,4,6}}Output
7
186 Views
