# Queries to check if any pair exists in an array having values at most equal to the given pair

• Difficulty Level : Expert
• Last Updated : 25 Jun, 2021

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++14

 // C++ program for the above approach#include using namespace std; // Function that performs binary search// to find value less than or equal to// first value of the pairint binary_search(vector > 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 pairvoid modify_vec(    vector >& 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 queryint evaluate_query(vector > 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 queryvoid checkPairs(vector >& v,                vector >& 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 Codeint main(){    vector > arr        = { { 3, 5 }, { 2, 7 }, { 2, 3 }, { 4, 9 } };     vector > queries        = { { 3, 4 }, { 3, 2 }, { 4, 1 }, { 3, 7 } };     // Function Call    checkPairs(arr, queries);     return 0;}

## Java

 // Java program for above approachimport 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

 # Python3 program for the above approach # Function that performs binary search# to find value less than or equal to# first value of the pairdef 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 pairdef 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 querydef 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 querydef 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 Codeif __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

## Javascript



Output:

Yes
No
No
Yes

Time Complexity: O(Q*log N)
Auxiliary Space: O(1)

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