Given a vector of pairs arr[] and Q queries in the form of pairs in an array Queries[], the task for each query is to check if there exists any pair with both the values smaller than those in the pair of the current query. If found to be true, then print “Yes”. Otherwise, print “No”.
Examples:
Input: arr[][] = {{3, 5}, {2, 7}, {2, 3}, {4, 9}}, Queries[][] = {{3, 4}, {3, 2}, {4, 1}, {3, 7}}
Output:
Yes
No
No
Yes
Explanation:
Query 1: Pair {2, 3} exists in arr[] which has both the values smaller than {3, 4}.
Query 2: No valid pair found in arr[] for {3, 2}.
Query 3: No valid pair found in arr[] for {4, 1}.
Query 4: Pair {2, 7} exists in arr[] for {3, 7}.Input: arr[][] = {{2, 1}, {4, 2}, {4, 4}, {7, 2}}, Queries[][] = {{2, 1}, {1, 1}}
Output:
Yes
No
Naive Approach: The simplest approach is to traverse the array Queries[][] and for each pair, traverse the given array of pairs and check if there exists any such pair whose corresponding values is greater than or equal to the pair {p1, p2} then print “Yes”. Otherwise, print “No”.
Time Complexity: O(N*K)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to use Binary Search. Follow the steps below to solve this problem:
- Sort the array of pairs with respect to first element in the pairs in increasing order. If there exist 2 pairs whose first values are the same, then pairs are arranged on the basis of the second element of the pair.
- After sorting, traverse the array of pairs, and for all pairs having the same first value, replace the second value with the minimum of all the pairs having the same first value.
- Now, traverse the given Queries[] arrays and perform binary search on the array arr[] for each pair in it.
- If the pairs obtained from the above steps is smaller than the current pair in the Query[], then print “Yes” else print “No”.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function that performs binary search // to find value less than or equal to // first value of the pair int binary_search(vector<pair< int , int > > vec,
int n, int a)
{ int low, high, mid;
low = 0;
high = n - 1;
// Perform binary search
while (low < high) {
// Find the mid
mid = low + (high - low + 1) / 2;
// Update the high
if (vec[mid].first > a) {
high = mid - 1;
}
// Else update low
else if (vec[mid].first <= a) {
low = mid;
}
}
// Return the low index
return low;
} // Function to modify the second // value of each pair void modify_vec(
vector<pair< int , int > >& v, int n)
{ // start from index 1
for ( int i = 1; i < n; i++) {
v[i].second
= min(v[i].second,
v[i - 1].second);
}
} // Function to evaluate each query int evaluate_query(vector<pair< int , int > > v,
int n, int m1,
int m2)
{ // Find value less than or equal to
// the first value of pair
int temp = binary_search(v, n, m1);
// check if we got the required
// pair or not
if ((v[temp].first <= m1)
&& (v[temp].second <= m2)) {
return 1;
}
return 0;
} // Function to find a pair whose values is // less than the given pairs in query void checkPairs(vector<pair< int , int > >& v,
vector<pair< int , int > >& queries)
{ // Find the size of the vector
int n = v.size();
// sort the vector based on
// the first value
sort(v.begin(), v.end());
// Function Call to modify the
// second value of each pair
modify_vec(v, n);
int k = queries.size();
// Traverse each queries
for ( int i = 0; i < k; i++) {
int m1 = queries[i].first;
int m2 = queries[i].second;
// Evaluate each query
int result
= evaluate_query(v, n, m1, m2);
// Print the result
if (result > 0)
cout << "Yes\n" ;
else
cout << "No\n" ;
}
} // Driver Code int main()
{ vector<pair< int , int > > arr
= { { 3, 5 }, { 2, 7 }, { 2, 3 }, { 4, 9 } };
vector<pair< int , int > > queries
= { { 3, 4 }, { 3, 2 }, { 4, 1 }, { 3, 7 } };
// Function Call
checkPairs(arr, queries);
return 0;
} |
// Java program for above approach import java.util.*;
import java.lang.*;
class GFG{
// Function that performs binary search
// to find value less than or equal to
// first value of the pair
static int binary_search( int [][] vec,
int n, int a)
{
int low, high, mid;
low = 0 ;
high = n - 1 ;
// Perform binary search
while (low < high)
{
// Find the mid
mid = low + (high - low + 1 ) / 2 ;
// Update the high
if (vec[mid][ 0 ] > a)
{
high = mid - 1 ;
}
// Else update low
else if (vec[mid][ 1 ] <= a)
{
low = mid;
}
}
// Return the low index
return low;
}
// Function to modify the second
// value of each pair
static void modify_vec(
int [][] v, int n)
{
// start from index 1
for ( int i = 1 ; i < n; i++)
{
v[i][ 1 ] = Math.min(v[i][ 1 ],
v[i - 1 ][ 1 ]);
}
}
// Function to evaluate each query
static int evaluate_query( int [][] v,
int n, int m1,
int m2)
{
// Find value less than or equal to
// the first value of pair
int temp = binary_search(v, n, m1);
// check if we got the required
// pair or not
if ((v[temp][ 0 ] <= m1)
&& (v[temp][ 1 ] <= m2))
{
return 1 ;
}
return 0 ;
}
// Function to find a pair whose values is
// less than the given pairs in query
static void checkPairs( int [][] v,
int [][] queries)
{
// Find the size of the vector
int n = v.length;
// sort the vector based on
// the first value
Arrays.sort(v, (a, b)->a[ 0 ]-b[ 0 ]);
// Function Call to modify the
// second value of each pair
modify_vec(v, n);
int k = queries.length;
// Traverse each queries
for ( int i = 0 ; i < k; i++)
{
int m1 = queries[i][ 0 ];
int m2 = queries[i][ 1 ];
// Evaluate each query
int result
= evaluate_query(v, n, m1, m2);
// Print the result
if (result > 0 )
System.out.println( "Yes" );
else
System.out.println( "No" );
}
}
// Driver function
public static void main (String[] args)
{
int [][] arr
= { { 3 , 5 }, { 2 , 7 }, { 2 , 3 }, { 4 , 9 } };
int [][] queries
= { { 3 , 4 }, { 3 , 2 }, { 4 , 1 }, { 3 , 7 } };
// Function Call
checkPairs(arr, queries);
}
} // This code is contributed by offbeat |
# Python3 program for the above approach # Function that performs binary search # to find value less than or equal to # first value of the pair def binary_search(vec, n, a):
low, high, mid = 0 , 0 , 0
low = 0
high = n - 1
# Perform binary search
while (low < high):
# Find the mid
mid = low + (high - low + 1 ) / / 2
# Update the high
if (vec[mid][ 0 ] > a):
high = mid - 1
# Else update low
elif vec[mid][ 0 ] < = a:
low = mid
# Return the low index
return low
# Function to modify the second # value of each pair def modify_vec(v, n):
# start from index 1
for i in range (n):
v[i][ 1 ] = min (v[i][ 1 ], v[i - 1 ][ 1 ])
return v
# Function to evaluate each query def evaluate_query(v, n, m1, m2):
# Find value less than or equal to
# the first value of pair
temp = binary_search(v, n, m1)
# check if we got the required
# pair or not
if ((v[temp][ 0 ] < = m1)
and (v[temp][ 1 ] < = m2)):
return 1
return 0
# Function to find a pair whose values is # less than the given pairs in query def checkPairs(v, queries):
# Find the size of the vector
n = len (v)
# sort the vector based on
# the first value
v = sorted (v)
# Function Call to modify the
# second value of each pair
v = modify_vec(v, n)
k = len (queries)
# Traverse each queries
for i in range (k):
m1 = queries[i][ 0 ]
m2 = queries[i][ 1 ]
# Evaluate each query
result = evaluate_query(v, n, m1, m2)
# Print the result
if (result > 0 ):
print ( "Yes" )
else :
print ( "No" )
# Driver Code if __name__ = = '__main__' :
arr = [ [ 3 , 5 ], [ 2 , 7 ], [ 2 , 3 ], [ 4 , 9 ] ]
queries = [ [ 3 , 4 ], [ 3 , 2 ], [ 4 , 1 ], [ 3 , 7 ] ]
# Function Call
checkPairs(arr, queries)
# This code is contributed by mohit kumar 29 |
// C# program for above approach using System;
using System.Collections.Generic;
class GFG{
// Function that performs binary search
// to find value less than or equal to
// first value of the pair
static int binary_search( int [][] vec,
int n, int a)
{
int low, high, mid;
low = 0;
high = n - 1;
// Perform binary search
while (low < high)
{
// Find the mid
mid = low + (high - low + 1) / 2;
// Update the high
if (vec[mid][0] > a)
{
high = mid - 1;
}
// Else update low
else if (vec[mid][1] <= a)
{
low = mid;
}
}
// Return the low index
return low;
}
// Function to modify the second
// value of each pair
static void modify_vec(
int [][] v, int n)
{
// start from index 1
for ( int i = 1; i < n; i++)
{
v[i][1] = Math.Min(v[i][1],
v[i - 1][1]);
}
}
// Function to evaluate each query
static int evaluate_query( int [][] v,
int n, int m1,
int m2)
{
// Find value less than or equal to
// the first value of pair
int temp = binary_search(v, n, m1);
// check if we got the required
// pair or not
if ((v[temp][0] <= m1)
&& (v[temp][1] <= m2))
{
return 1;
}
return 0;
}
static int compare( int [] a, int [] b)
{
if (a[0] < b[0])
return -1;
if (a[0] == b[0])
return 0;
return 1;
}
// Function to find a pair whose values is
// less than the given pairs in query
static void checkPairs( int [][] v,
int [][] queries)
{
// Find the size of the vector
int n = v.Length;
// sort the vector based on
// the first value
Array.Sort(v, compare);
// Function Call to modify the
// second value of each pair
modify_vec(v, n);
int k = queries.Length;
// Traverse each queries
for ( int i = 0; i < k; i++)
{
int m1 = queries[i][0];
int m2 = queries[i][1];
// Evaluate each query
int result
= evaluate_query(v, n, m1, m2);
// Print the result
if (result > 0)
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
}
// Driver function
public static void Main ( string [] args)
{
int [][] arr
= { new int [] { 3, 5 }, new int [] { 2, 7 }, new int [] { 2, 3 }, new int [] { 4, 9 } };
int [][] queries
= { new int [] { 3, 4 }, new int [] { 3, 2 }, new int [] { 4, 1 }, new int [] { 3, 7 } };
// Function Call
checkPairs(arr, queries);
}
} // This code is contributed by phasing17 |
<script> // Javascript program for above approach // Function that performs binary search // to find value less than or equal to
// first value of the pair
function binary_search(vec,a,n)
{ let low, high, mid;
low = 0;
high = n - 1;
// Perform binary search
while (low < high)
{
// Find the mid
mid = low + Math.floor((high - low + 1) / 2);
// Update the high
if (vec[mid][0] > a)
{
high = mid - 1;
}
// Else update low
else if (vec[mid][1] <= a)
{
low = mid;
}
}
// Return the low index
return low;
} // Function to modify the second
// value of each pair
function modify_vec(v,n)
{ // start from index 1
for (let i = 1; i < n; i++)
{
v[i][1] = Math.min(v[i][1],
v[i - 1][1]);
}
} // Function to evaluate each query function evaluate_query(v,n,m1,m2)
{ // Find value less than or equal to
// the first value of pair
let temp = binary_search(v, n, m1);
// check if we got the required
// pair or not
if ((v[temp][0] <= m1)
&& (v[temp][1] <= m2))
{
return 1;
}
return 0;
} // Function to find a pair whose values is
// less than the given pairs in query
function checkPairs(v, queries)
{ // Find the size of the vector
let n = v.length;
// sort the vector based on
// the first value
v.sort( function (a, b){ return a[0]-b[0]});
// Function Call to modify the
// second value of each pair
modify_vec(v, n);
let k = queries.length;
// Traverse each queries
for (let i = 0; i < k; i++)
{
let m1 = queries[i][0];
let m2 = queries[i][1];
// Evaluate each query
let result
= evaluate_query(v, n, m1, m2);
// Print the result
if (result > 0)
document.write( "Yes<br>" );
else
document.write( "No<br>" );
}
} // Driver function let arr=[[ 3, 5 ], [ 2, 7 ], [ 2, 3 ], [ 4, 9 ]]; let queries=[[ 3, 4 ], [ 3, 2 ], [ 4, 1 ], [ 3, 7 ]]; // Function Call checkPairs(arr, queries); // This code is contributed by unknown2108 </script> |
Yes No No Yes
Time Complexity: O(Q*log N)
Auxiliary Space: O(1)