# Find smallest range containing elements from k lists in C++

C++Server Side ProgrammingProgramming

Suppose we have k different lists. The elements are sorted. We have to search the smallest range that includes at least one number from each of the k different 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

## Example

Let us see the following implementation to get better understanding −

Live Demo

#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, ]
Published on 19-Aug-2020 11:21:41