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Smallest Range Covering Elements from K Lists in C++
Suppose we have k lists of sorted integers. We have to search the smallest range that includes at least one number from each of the k lists. Here the range [a,b] is smaller than range [c,d] when b-a < d-c or a < c if b-a == d-c.
So if the input is like [[4,10,15,25,26],[0,9,14,20],[5,18,24,30]], then the output will be [14, 18]
To solve this, we will follow these steps −
- minRange := inf, maxRange := -inf, rangeSize := inf, tempMinRange := inf, tempMaxRange := -inf
- n := size of nums
- Define an array pointers of size n
- make a priority queue pq
- for initialize i := 0, when i < n, update (increase i by 1), do −
- insert { nums[i, 0], i } into pq
- tempMaxRange := maximum of tempMaxRange and nums[i, 0]
- while 1 is non-zero, do −
- Define one pair temp := top of pq
- delete element from pq
- tempMinRange := temp.first
- idx := second element of temp
- if tempMaxRange − tempMinRange < rangeSize, then −
- rangeSize := tempMaxRange - tempMinRange
- minRange := tempMinRange
- maxRange := tempMaxRange
- (increase pointers[idx] by 1)
- if pointers[idx] is same as size of nums[idx], then −
- Come out from the loop
- Otherwise
- tempMaxRange := maximum of tempMaxRange and nums[idx, pointers[idx]]
- insert { nums[idx, pointers[idx]], idx } into pq
- Define an array ans of size 2
- ans[0] := minRange, ans[1] := maxRange
- return ans
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
struct Comparator{
bool operator() (pair <int, int> a, pair <int, int> b){
return !(a.first < b.first);
}
};
class Solution {
public:
vector<int> smallestRange(vector<vector<int>>& nums) {
int minRange = INT_MAX;
int maxRange = INT_MIN;
int rangeSize = INT_MAX;
int tempMinRange, tempMaxRange, tempRangeSize;
tempMinRange = INT_MAX;
tempMaxRange = INT_MIN;
int n = nums.size();
vector <int> pointers(n);
priority_queue < pair <int, int>, vector < pair <int, int> >, Comparator > pq;
for(int i = 0; i < n; i++){
pq.push({nums[i][0], i});
tempMaxRange = max(tempMaxRange, nums[i][0]);
}
while(1){
pair <int, int> temp = pq.top();
pq.pop();
tempMinRange = temp.first;
int idx = temp.second;
if(tempMaxRange - tempMinRange < rangeSize){
rangeSize = tempMaxRange - tempMinRange;
minRange = tempMinRange;
maxRange = tempMaxRange;
}
pointers[idx]++;
if(pointers[idx] == nums[idx].size())break;
else{
tempMaxRange = max(tempMaxRange, nums[idx][pointers[idx]]);
pq.push({nums[idx][pointers[idx]], idx});
}
}
vector <int> ans(2);
ans[0] = minRange;
ans[1] = maxRange;
return ans;
}
};
main(){
Solution ob;
vector<vector<int>> v = {{4,10,15,25,26},{0,9,14,20},{5,18,24,30}};
print_vector(ob.smallestRange(v));
}Input
{{4,10,15,25,26},{0,9,14,20},{5,18,24,30}};Output
[14, 18, ]
Updated on: 01-Jun-2020
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