Given N ranges of the form [L, R], the task is to find the range having the minimum number of integers such that at least one point of all the given N ranges exists in that range.
Example:
Input: ranges[] = {{1, 6}, {2, 7}, {3, 8}, {4, 9}}
Output: 6 6
Explanation: All the given ranges contain 6 as an integer between them. Therefore, [6, 6] is a valid range having 1 integer which is the minimum possible. The other valid ranges are [4, 4] and [5, 5].Input: ranges[] = {{1, 4}, {4, 5}, {7, 9}, {9, 12}}
Output: 4 9
Approach: This problem can be solved using a Greedy Approach using the following observations:
- Suppose L is the minimum endpoint over all the given ranges. Then, it can be observed that all the given ranges must contain a point in the range [L, ∞].
- Similarly, suppose R is the maximum starting point over all the given ranges. Then, based on the similar observation as above, all the ranges must also contain a point in the range [-∞, R].
Therefore, using the above observations, the required range will be [L, R]. In cases where L > R, the answer can be any integer between L and R as all of them will make a valid range.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find minimum range such // that alteast one point of all the // given N ranges exist in the range void minRequiredRange(
vector<pair< int , int > > ranges,
int N)
{ // Stores the starting point of
// the required range
int L = INT_MAX;
// Stores the ending point of the
// required range
int R = INT_MIN;
// Loop to iterate over all the given
// N ranges
for ( int i = 0; i < N; i++) {
// Calculate the smallest end point
// over all the given ranges
L = min(L, ranges[i].second);
// Calculate the largest starting point
// over all the given ranges
R = max(R, ranges[i].first);
}
// If starting point is greater that the
// end point
if (R < L) {
// Print any integer between L and R
cout << L << " " << L;
}
else {
// Print Answer
cout << L << " " << R;
}
} // Driver Code int main()
{ vector<pair< int , int > > ranges{
{ 1, 4 }, { 4, 5 }, { 7, 9 }, { 9, 12 }
};
minRequiredRange(ranges, ranges.size());
return 0;
} |
// Java program for the above approach class GFG {
// Function to find minimum range such
// that alteast one point of all the
// given N ranges exist in the range
public static void minRequiredRange( int [][] ranges, int N) {
// Stores the starting point of
// the required range
int L = Integer.MAX_VALUE;
// Stores the ending point of the
// required range
int R = Integer.MIN_VALUE;
// Loop to iterate over all the given
// N ranges
for ( int i = 0 ; i < N; i++) {
// Calculate the smallest end point
// over all the given ranges
L = Math.min(L, ranges[i][ 1 ]);
// Calculate the largest starting point
// over all the given ranges
R = Math.max(R, ranges[i][ 0 ]);
}
// If starting point is greater that the
// end point
if (R < L) {
// Print any integer between L and R
System.out.println(L + " " + L);
} else {
// Print Answer
System.out.println(L + " " + R);
}
}
// Driver Code
public static void main(String args[]) {
int [][] ranges = { { 1 , 4 }, { 4 , 5 }, { 7 , 9 }, { 9 , 12 } };
minRequiredRange(ranges, ranges.length);
}
} // This code is contributed by gfgking. |
# Python Program to implement # the above approach # Function to find minimum range such # that alteast one point of all the # given N ranges exist in the range def minRequiredRange(ranges, N):
# Stores the starting point of
# the required range
L = 10 * * 9
# Stores the ending point of the
# required range
R = 10 * * - 9
# Loop to iterate over all the given
# N ranges
for i in range (N):
# Calculate the smallest end point
# over all the given ranges
L = min (L, ranges[i][ "second" ])
# Calculate the largest starting point
# over all the given ranges
R = max (R, ranges[i][ "first" ])
# If starting point is greater that the
# end point
if (R < L):
# Print any integer between L and R
print (f "{L} {L}" )
else :
# Print Answer
print (f "{L} {R}" )
# Driver Code ranges = [{ "first" : 1 , "second" : 4 }, { "first" : 4 , "second" : 5 }, {
"first" : 7 , "second" : 9 }, { "first" : 9 , "second" : 12 }
] minRequiredRange(ranges, len (ranges))
# This code is contributed by gfgking |
// C# program for the above approach using System;
class GFG {
// Function to find minimum range such
// that alteast one point of all the
// given N ranges exist in the range
public static void minRequiredRange( int [, ] ranges,
int N)
{
// Stores the starting point of
// the required range
int L = Int32.MaxValue;
// Stores the ending point of the
// required range
int R = Int32.MinValue;
// Loop to iterate over all the given
// N ranges
for ( int i = 0; i < N; i++) {
// Calculate the smallest end point
// over all the given ranges
L = Math.Min(L, ranges[i, 1]);
// Calculate the largest starting point
// over all the given ranges
R = Math.Max(R, ranges[i, 0]);
}
// If starting point is greater that the
// end point
if (R < L) {
// Print any integer between L and R
Console.WriteLine(L + " " + L);
}
else {
// Print Answer
Console.WriteLine(L + " " + R);
}
}
// Driver Code
public static void Main( string [] args)
{
int [, ] ranges
= { { 1, 4 }, { 4, 5 }, { 7, 9 }, { 9, 12 } };
minRequiredRange(ranges, ranges.GetLength(0));
}
} // This code is contributed by ukasp. |
<script> // JavaScript Program to implement
// the above approach
// Function to find minimum range such
// that alteast one point of all the
// given N ranges exist in the range
function minRequiredRange(
ranges,
N) {
// Stores the starting point of
// the required range
let L = Number.MAX_VALUE;
// Stores the ending point of the
// required range
let R = Number.MIN_VALUE;
// Loop to iterate over all the given
// N ranges
for (let i = 0; i < N; i++) {
// Calculate the smallest end point
// over all the given ranges
L = Math.min(L, ranges[i].second);
// Calculate the largest starting point
// over all the given ranges
R = Math.max(R, ranges[i].first);
}
// If starting point is greater that the
// end point
if (R < L) {
// Print any integer between L and R
document.write(L + " " + L);
}
else {
// Print Answer
document.write(L + " " + R);
}
}
// Driver Code
let ranges = [
{ first: 1, second: 4 }, { first: 4, second: 5 }, { first: 7, second: 9 }, { first: 9, second: 12 }
];
minRequiredRange(ranges, ranges.length);
// This code is contributed by Potta Lokesh
</script>
|
4 9
Time Complexity: O(N)
Auxiliary Space: O(1)