Queries to count array elements greater than or equal to a given number with updates

Namachila November 19, 2025

0 1 15 minutes read

Given two arrays arr[] and query[] of sizes N and Q respectively and an integer M, the task for each query, is to count the number of array elements that are greater than or equal to query[i] and decrease all such numbers by M and perform the rest of the queries on the updated array.

Examples:

Input: arr[] = {1, 2, 3, 4}, query[] = {4, 3, 1}, M = 1
Output: 1 2 4
Explanation:
query[0]: Count of array elements which are greater than or equal to 4 in arr[] is 1 and decreasing the number by M modifies array to {1, 2, 3, 3}.
query[1]: Count of array elements which are greater than or equal to 3 in arr[] is 2 and decreasing all such numbers by M modifies array to {1, 2, 2, 2}.
query[2]: Count of array elements which are greater than or equal to 1 in arr[] is 4 and decreasing all such numbers by M modifies array to {0, 1, 1, 1}.

Input: arr[] = {1, 2, 3}, query = {3, 3}, M = 2
Output: 1 0
Explanation:
query[0]: Count of array elements which are greater than or equal to in arr[] is 1 and decreasing that number by M modifies array to arr[] = {1, 2, 1}.
query[1]: Count of array elements which are greater than or equal to 3 in arr[] is 0. So the array remains unchanged.

Naive Approach: The simplest approach is to iterate over the entire array for every query, and check if the current array element is greater than or equal to query[i]. For elements for which the above condition is found to be true, subtract that element by M and increment the count. Finally, print the count for each query.

Below is the implementation of the above approach:

C++14

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the numbers
// which are greater than the given query
void sequenceMaintenance(int arr[], int n, int query[],
                         int q, int m)
{
 
    // Iterate over the query array
    for (int i = 0; i < q; i++) {
 
        int count = 0;
 
        // Iterate over the input
        // array for each query
        for (int j = 0; j < n; j++) {
 
            // curr element greater than the
            // given query increment the count
            // and decrement the element by m
            if (arr[j] >= query[i]) {
                arr[j] -= m;
                count++;
            }
        }
 
        cout << count << " ";
    }
}
 
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int query[] = { 4, 3, 1 };
    int m = 1;
    int n = sizeof(arr) / sizeof(arr[0]);
    int q = sizeof(query) / sizeof(query[0]);
 
    // Function Call
    sequenceMaintenance(arr, n, query, q, m);
 
    return 0;
}

Output:

1 2 4

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

Efficient Approach: The problem can be solved using Segment Trees.

First, sort the array arr[] in non decreasing order.
Now, find the first position of element, say l, which contains an element >= query[i].
If no such elements exist then answer for that query will be 0. Otherwise answer will be N – l.
Finally update the segment tree in the given range l to N – 1 and subtract M from all elements in the given range.

Illustration:
Consider the following example : arr[] = {1, 2, 3, 4}, query[] = {4, 3, 1}, M = 1
After sorting arr[] = {1, 2, 3, 4}.
query[0]: K = 4 and arr[3] >= 4 so l = 3 and result = 4 – 3 = 1 and updated arr[] = {1, 2, 3, 3}
query[1]: K = 3 and arr[2] >=3 so l = 2 and result = 4 – 2 = 2 and updated arr[] = {1, 2, 2, 2}
query[2]: K = 1 and arr[0] >=1 so l = 0 and result = 4 – 0 = 4 and updated arr[] = {0, 1, 1, 1}

Below is the implementation of above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to build a segment tree
void build(vector<int>& sum,
           vector<int>& a,
           int l, int r, int rt)
{
    // Check for base case
    if (l == r) {
        sum[rt] = a[l - 1];
        return;
    }
 
    // Find mid point
    int m = (l + r) >> 1;
 
    // Recursively build the
    // segment tree
    build(sum, a, l, m, rt << 1);
    build(sum, a, m + 1, r, rt << 1 | 1);
}
 
// Function for push down operation
// on the segment tree
void pushDown(vector<int>& sum,
              vector<int>& add,
              int rt, int ln, int rn)
{
    if (add[rt]) {
        add[rt << 1] += add[rt];
        add[rt << 1 | 1] += add[rt];
        sum[rt << 1] += add[rt] * ln;
        sum[rt << 1 | 1] += add[rt] * rn;
        add[rt] = 0;
    }
}
 
// Function to update the segment tree
void update(vector<int>& sum,
            vector<int>& add,
            int L, int R, int C, int l,
            int r, int rt)
{
    // Complete overlap
    if (L <= l && r <= R) {
        sum[rt] += C * (r - l + 1);
        add[rt] += C;
        return;
    }
 
    // Find mid
    int m = (l + r) >> 1;
 
    // Perform push down operation
    // on segment tree
    pushDown(sum, add, rt, m - l + 1,
             r - m);
 
    // Recursively update the segment tree
    if (L <= m)
        update(sum, add, L, R, C, l, m,
               rt << 1);
 
    if (R > m)
        update(sum, add, L, R, C, m + 1, r,
               rt << 1 | 1);
}
 
// Function to process the query
int query(vector<int>& sum, 
          vector<int>& add,
          int L, int R, int l, 
          int r, int rt)
{
    // Base case
    if (L <= l && r <= R) {
        return sum[rt];
    }
 
    // Find mid
    int m = (l + r) >> 1;
 
    // Perform push down operation
    // on segment tree
    pushDown(sum, add, rt, m - l + 1,
             r - m);
 
    int ans = 0;
 
    // Recursively calculate the result
    // of the query
    if (L <= m)
        ans += query(sum, add, L, R, l, m,
                     rt << 1);
    if (R > m)
        ans += query(sum, add, L, R, m + 1, r,
                     rt << 1 | 1);
 
    // Return the result
    return ans;
}
 
// Function to count the numbers
// which are greater than the given query
void sequenceMaintenance(int n, int q,
                         vector<int>& a,
                         vector<int>& b, 
                         int m)
{
    // Sort the input array
    sort(a.begin(), a.end());
 
    // Create segment tree of size 4*n
    vector<int> sum, add, ans;
    sum.assign(n << 2, 0);
    add.assign(n << 2, 0);
 
    // Build the segment tree
    build(sum, a, 1, n, 1);
 
    // Iterate over the queries
    for (int i = 0; i < q; i++) {
        int l = 1, r = n, pos = -1;
        while (l <= r) {
            int m = (l + r) >> 1;
            if (query(sum, add, m, m, 1, n, 1)
                >= b[i]) {
                r = m - 1;
                pos = m;
            }
            else {
                l = m + 1;
            }
        }
        if (pos == -1)
            ans.push_back(0);
        else {
            // Store result in array
            ans.push_back(n - pos + 1);
 
            // Update the elements in
            // the given range
            update(sum, add, pos, n, -m,
                   1, n, 1);
        }
    }
 
    // Print the result of queries
    for (int i = 0; i < ans.size(); i++) {
        cout << ans[i] << " ";
    }
}
 
// Driver Code
int main()
{
    int N = 4;
    int Q = 3;
    int M = 1;
    vector<int> arr = { 1, 2, 3, 4 };
    vector<int> query = { 4, 3, 1 };
 
    // Function Call
    sequenceMaintenance(N, Q, arr, query, M);
    return 0;
}

Java

// Java program for the above approach 
import java.io.*;
import java.util.*;
class GFG 
{
   
    // Function to build a segment tree 
    static void build(Vector<Integer> sum,Vector<Integer> a,
                      int l, int r, int rt)
    {
       
        // Check for base case 
        if(l == r)
        {
            sum.set(rt, a.get(l - 1));
            return;
        }
       
        // Find mid point 
        int m = (l + r) >> 1; 
       
        // Recursively build the 
        // segment tree 
        build(sum, a, l, m, rt << 1); 
        build(sum, a, m + 1, r, rt << 1 | 1);
         
    }
   
    // Function for push down operation 
    // on the segment tree 
    static void pushDown(Vector<Integer> sum, 
                         Vector<Integer> add,
                         int rt, int ln, int rn)
    {
        if(add.get(rt) != 0)
        {
            add.set(rt << 1, add.get(rt));
            add.set(rt << 1 | 1, add.get(rt));
            sum.set(rt << 1, sum.get(rt << 1) + add.get(rt) * ln);
            sum.set(rt << 1 | 1, sum.get(rt << 1 | 1) + add.get(rt) * rn);
            add.set(rt, 0);
        }
    }
   
    // Function to update the segment tree
    static void update(Vector<Integer> sum, 
                       Vector<Integer> add,int L, 
                       int R, int C, int l, int r, int rt)
    {
       
        // Complete overlap 
        if(L <= l && r <= R)
        {
            sum.set(rt,sum.get(rt) + C * (r - l + 1));
            add.set(rt,add.get(rt) + C);
            return;
        }
       
        // Find mid 
        int m = (l + r) >> 1; 
       
        // Perform push down operation 
        // on segment tree
        pushDown(sum, add, rt, m - l + 1, r - m);
       
        // Recursively update the segment tree 
        if(L <= m)
        {
            update(sum, add, L, R, C, l, m, rt << 1);
             
        }
        if(R > m)
        {
            update(sum, add, L, R, C, m + 1, r, rt << 1 | 1);
        }
    }
    // Function to process the query
    static int query(Vector<Integer> sum,Vector<Integer> add,
                     int L, int R, int l,int r, int rt)
    {
        // Base case 
        if (L <= l && r <= R)
        {
            return sum.get(rt);
        }
       
        // Find mid
        int m = (l + r) >> 1;
       
        // Perform push down operation 
        // on segment tree
        pushDown(sum, add, rt, m - l + 1, r - m); 
        int ans = 0;
       
        // Recursively calculate the result 
        // of the query 
        if(L <= m)
        {
            ans += query(sum, add, L, R, l, m, rt << 1); 
             
        }
        if(R > m)
        {
            ans += query(sum, add, L, R, m + 1, r,rt << 1 | 1); 
        }
       
        // Return the result
        return ans;
    }
   
    // Function to count the numbers 
    // which are greater than the given query 
    static void sequenceMaintenance(int n, int q,
                                    Vector<Integer> a,
                                    Vector<Integer> b,int m)
    {
        // Sort the input array
        Collections.sort(a);
       
        // Create segment tree of size 4*n
        Vector<Integer> sum = new Vector<Integer>();
        Vector<Integer> ad = new Vector<Integer>();
        Vector<Integer> ans = new Vector<Integer>();
        for(int i = 0; i < (n << 2); i++)
        {
            sum.add(0);
            ad.add(0);
        }
       
        // Build the segment tree 
        build(sum, a, 1, n, 1); 
       
        // Iterate over the queries 
        for(int i = 0; i < q; i++)
        {
            int l = 1, r = n, pos = -1; 
            while(l <= r)
            {
                m = (l + r) >> 1;
                if(query(sum, ad, m, m, 1, n, 1) >= b.get(i))
                {
                    r = m - 1; 
                    pos = m;
                }
                else
                {
                    l = m + 1; 
                }
            }
            if(pos == -1)
            {
                ans.add(0);
            }
            else
            {
               
                // Store result in array 
                ans.add(n - pos + 1);
               
                // Update the elements in 
                // the given range 
                update(sum, ad, pos, n, -m, 1, n, 1);
            }
        }
       
         // Print the result of queries 
        for(int i = 0; i < ans.size(); i++)
        {
            System.out.print(ans.get(i) + " ");
        }
    }
   
    // Driver Code
    public static void main (String[] args) 
    {
        int N = 4; 
        int Q = 3; 
        int M = 1;
        Vector<Integer> arr = new Vector<Integer>();
        arr.add(1);
        arr.add(2);
        arr.add(3);
        arr.add(4);
        Vector<Integer> query = new Vector<Integer>();
        query.add(4);
        query.add(3);
        query.add(1);
       
        // Function call 
        sequenceMaintenance(N, Q, arr, query, M);
    }
}
 
// This code is contributed by rag2127

Python3

# Python3 program for the above approach
 
# Function to build a segment tree
def build(sum, a, l, r, rt):
 
    # Check for base case
    if (l == r):
        sum[rt] = a[l - 1]
        return
         
    # Find mid point
    m = (l + r) >> 1
 
    # Recursively build the
    # segment tree
    build(sum, a, l, m, rt << 1)
    build(sum, a, m + 1, r, rt << 1 | 1)
 
# Function for push down operation
# on the segment tree
def pushDown(sum, add, rt, ln, rn):
 
    if (add[rt]):
        add[rt << 1] += add[rt]
        add[rt << 1 | 1] += add[rt]
        sum[rt << 1] += add[rt] * ln
        sum[rt << 1 | 1] += add[rt] * rn
        add[rt] = 0
 
# Function to update the segment tree
def update(sum, add, L, R, C, l, r, rt):
 
    # Complete overlap
    if (L <= l and r <= R):
        sum[rt] += C * (r - l + 1)
        add[rt] += C
        return
 
    # Find mid
    m = (l + r) >> 1
 
    # Perform push down operation
    # on segment tree
    pushDown(sum, add, rt, m - l + 1, r - m)
 
    # Recursively update the segment tree
    if (L <= m):
        update(sum, add, L, R, C, l,
               m, rt << 1)
 
    if (R > m):
        update(sum, add, L, R, C, m + 1,
               r, rt << 1 | 1)
 
# Function to process the queryy
def queryy(sum, add, L, R, l, r, rt):
 
    # Base case
    if (L <= l and r <= R):
        return sum[rt]
 
    # Find mid
    m = (l + r) >> 1
 
    # Perform push down operation
    # on segment tree
    pushDown(sum, add, rt, m - l + 1, r - m)
 
    ans = 0
 
    # Recursively calculate the result
    # of the queryy
    if (L <= m):
        ans += queryy(sum, add, L, R, l, 
                      m, rt << 1)
    if (R > m):
        ans += queryy(sum, add, L, R, m + 1,
                      r, (rt << 1 | 1))
 
    # Return the result
    return ans
 
# Function to count the numbers
# which are greater than the given queryy
def sequenceMaintenance(n, q, a, b, m):
 
    # Sort the input array
    a = sorted(a)
 
    # Create segment tree of size 4*n
    # vector<int> sum, add, ans
    sum = [0] * (4 * n)
    add = [0] * (4 * n)
    ans = []
 
    # Build the segment tree
    build(sum, a, 1, n, 1)
 
    #print(sum)
 
    # Iterate over the queries
    for i in range(q):
        l = 1
        r = n
        pos = -1
         
        while (l <= r):
            m = (l + r) >> 1
            if (queryy(sum, add, m, m,
                       1, n, 1) >= b[i]):
                r = m - 1
                pos = m
 
            else:
                l = m + 1
 
        if (pos == -1):
            ans.append(0)
        else:
             
            # Store result in array
            ans.append(n - pos + 1)
 
            # Update the elements in
            # the given range
            update(sum, add, pos, n, 
                   -m, 1, n, 1)
 
    # Print the result of queries
    for i in ans:
        print(i, end = " ")
 
# Driver Code
if __name__ == '__main__':
 
    N = 4
    Q = 3
    M = 1
     
    arr = [ 1, 2, 3, 4 ]
    query = [ 4, 3, 1 ]
 
    # Function call
    sequenceMaintenance(N, Q, arr, query, M)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach 
using System; 
using System.Collections;
using System.Collections.Generic;  
  
class GFG{ 
    
// Function to build a segment tree
static void build(ArrayList sum,
                  ArrayList a,
                  int l, int r, 
                  int rt)
{
     
    // Check for base case
    if (l == r)
    {
        sum[rt] = a[l - 1];
        return;
    }
   
    // Find mid point
    int m = (l + r) >> 1;
   
    // Recursively build the
    // segment tree
    build(sum, a, l, m, rt << 1);
    build(sum, a, m + 1, r, rt << 1 | 1);
}
   
// Function for push down operation
// on the segment tree
static void pushDown(ArrayList sum,
                     ArrayList add,
                     int rt, int ln, 
                     int rn)
{
    if ((int)add[rt] != 0) 
    {
        add[rt << 1] = (int)add[rt << 1] +
                       (int)add[rt];
        add[rt << 1 | 1] = (int)add[rt << 1 | 1] +
                           (int)add[rt];
        sum[rt << 1] = (int)sum[rt << 1] +
                       (int)add[rt] * ln;
        sum[rt << 1 | 1] = (int)sum[rt << 1 | 1] +
                           (int)add[rt] * rn;
        add[rt] = 0;
    }
}
   
// Function to update the segment tree
static void update(ArrayList sum,
                   ArrayList add,
                   int L, int R, int C,
                   int l, int r, int rt)
{
     
    // Complete overlap
    if (L <= l && r <= R)
    {
        sum[rt] = (int)sum[rt] + 
                    C * (r - l + 1);
        add[rt] = (int)add[rt] + C;
        return;
    }
   
    // Find mid
    int m = (l + r) >> 1;
   
    // Perform push down operation
    // on segment tree
    pushDown(sum, add, rt, m - l + 1,
                           r - m);
   
    // Recursively update the segment tree
    if (L <= m)
        update(sum, add, L, R, C, l, m,
               rt << 1);
    if (R > m)
        update(sum, add, L, R, C, m + 1, r,
               rt << 1 | 1);
}
   
// Function to process the query
static int query(ArrayList sum, 
                 ArrayList add,
                 int L, int R, int l, 
                 int r, int rt)
{
     
    // Base case
    if (L <= l && r <= R)
    {
        return (int)sum[rt];
    }
   
    // Find mid
    int m = (l + r) >> 1;
   
    // Perform push down operation
    // on segment tree
    pushDown(sum, add, rt, m - l + 1,
                           r - m);
   
    int ans = 0;
   
    // Recursively calculate the result
    // of the query
    if (L <= m)
        ans += query(sum, add, L, R, l, m,
                     rt << 1);
    if (R > m)
        ans += query(sum, add, L, R, m + 1, r,
                     rt << 1 | 1);
   
    // Return the result
    return ans;
}
   
// Function to count the numbers
// which are greater than the given query
static void sequenceMaintenance(int n, int q,
                                ArrayList a,
                                ArrayList b, 
                                int m)
{
     
    // Sort the input array
    a.Sort();
   
    // Create segment tree of size 4*n
    ArrayList sum = new ArrayList();
    ArrayList add = new ArrayList();
    ArrayList ans = new ArrayList();
      
    for(int i = 0; i < (n << 2); i++)
    {
        sum.Add(0);
        add.Add(0);
    }
      
    // Build the segment tree
    build(sum, a, 1, n, 1);
   
    // Iterate over the queries
    for(int i = 0; i < q; i++)
    {
        int l = 1, r = n, pos = -1;
        while (l <= r)
        {
            m = (l + r) >> 1;
            if (query(sum, add, m, m,
                      1, n, 1) >= (int)b[i])
            {
                r = m - 1;
                pos = m;
            }
            else
            {
                l = m + 1;
            }
        }
        if (pos == -1)
            ans.Add(0);
        else
        {
             
            // Store result in array
            ans.Add(n - pos + 1);
   
            // Update the elements in
            // the given range
            update(sum, add, pos, n, -m,
                   1, n, 1);
        }
    }
   
    // Print the result of queries
    for(int i = 0; i < ans.Count; i++)
    {
        Console.Write(ans[i] + " ");
    }
}
  
// Driver Code 
public static void Main(string[] args) 
{ 
    int N = 4;
    int Q = 3;
    int M = 1;
     
    ArrayList arr = new ArrayList(){ 1, 2, 3, 4 };
    ArrayList query = new ArrayList(){ 4, 3, 1 };
   
    // Function call
    sequenceMaintenance(N, Q, arr, query, M);
} 
} 
 
// This code is contributed by rutvik_56

Javascript

<script>
// Javascript program for the above approach
 
// Function to build a segment tree
function build(sum, a, l, r, rt)
{
 
    // Check for base case
        if(l == r)
        {
            sum[rt] =  a[l - 1];
            return;
        }
        
        // Find mid point
        let m = (l + r) >> 1;
        
        // Recursively build the
        // segment tree
        build(sum, a, l, m, rt << 1);
        build(sum, a, m + 1, r, rt << 1 | 1);
}
 
// Function for push down operation
    // on the segment tree
function pushDown(sum,add,rt,ln,rn)
{
    if(add[rt] != 0)
        {
            add[rt << 1] = add[rt];
            add[rt << 1 | 1] = add[rt];
            sum[rt << 1] = sum[rt << 1] + add[rt] * ln;
            sum[rt << 1 | 1] = sum[rt << 1 | 1] + add[rt] * rn;
            add[rt]= 0;
        }
}
 
// Function to update the segment tree
function update(sum,add,L,R,C,l,r,rt)
{
    // Complete overlap
        if(L <= l && r <= R)
        {
            sum[rt] = sum[rt] + C * (r - l + 1);
            add[rt] = add[rt] + C;
            return;
        }
        
        // Find mid
        let m = (l + r) >> 1;
        
        // Perform push down operation
        // on segment tree
        pushDown(sum, add, rt, m - l + 1, r - m);
        
        // Recursively update the segment tree
        if(L <= m)
        {
            update(sum, add, L, R, C, l, m, rt << 1);
              
        }
        if(R > m)
        {
            update(sum, add, L, R, C, m + 1, r, rt << 1 | 1);
        }
}
 
// Function to process the query
function query(sum,add,L,R,l,r,rt)
{
    // Base case
        if (L <= l && r <= R)
        {
            return sum[rt];
        }
        
        // Find mid
        let m = (l + r) >> 1;
        
        // Perform push down operation
        // on segment tree
        pushDown(sum, add, rt, m - l + 1, r - m);
        let ans = 0;
        
        // Recursively calculate the result
        // of the query
        if(L <= m)
        {
            ans += query(sum, add, L, R, l, m, rt << 1);
              
        }
        if(R > m)
        {
            ans += query(sum, add, L, R, m + 1, r,rt << 1 | 1);
        }
        
        // Return the result
        return ans;
}
 
// Function to count the numbers
    // which are greater than the given query
function sequenceMaintenance(n,q,a,b,m)
{
    // Sort the input array
        a.sort(function(a,b){return a-b;});
        
        // Create segment tree of size 4*n
        let sum = [];
        let ad = [];
        let ans = [];
        for(let i = 0; i < (n << 2); i++)
        {
            sum.push(0);
            ad.push(0);
        }
        
        // Build the segment tree
        build(sum, a, 1, n, 1);
        
        // Iterate over the queries
        for(let i = 0; i < q; i++)
        {
            let l = 1, r = n, pos = -1;
            while(l <= r)
            {
                m = (l + r) >> 1;
                if(query(sum, ad, m, m, 1, n, 1) >= b[i])
                {
                    r = m - 1;
                    pos = m;
                }
                else
                {
                    l = m + 1;
                }
            }
            if(pos == -1)
            {
                ans.push(0);
            }
            else
            {
                
                // Store result in array
                ans.push(n - pos + 1);
                
                // Update the elements in
                // the given range
                update(sum, ad, pos, n, -m, 1, n, 1);
            }
        }
        
         // Print the result of queries
        for(let i = 0; i < ans.length; i++)
        {
            document.write(ans[i] + " ");
        }
}
 
// Driver Code
let N = 4;
let Q = 3;
let M = 1;
 
let arr=[1, 2, 3, 4];
let Query = [4, 3, 1];
 
// Function call
sequenceMaintenance(N, Q, arr, Query, M);
 
// This code is contributed by patel2127
</script>

Output:

1 2 4

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

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!