Queries to find the first array element exceeding K with updates

Given an array arr[] of size N and a 2D array Q[][] consisting of queries of the following two types:

• 1 X Y: Update the array element at index X with Y.
• 2 K: Print the position of the first array element greater than or equal to K. If there is no such index, then print -1.

Examples:

Input : arr[] = { 1, 3, 2, 4, 6 }, Q[][] = {{2, 5}, {1, 3, 5}, {2, 4}, {2, 8}}
Output: 5 3 -1 
Explanation:
Query1: Since arr[4] > 5, the position of arr[4] is 5. 
Query2: Updating arr[2] with 5 modifies arr[] to {1, 3, 5, 4, 6} 
Query3: Since arr[2] > 4, the position of arr[4] is 5. 
Query4: No array element is greater than 8.

Input : arr[] = {1, 2, 3}, N = 3, Q[][] = {{2, 2}, {1, 3, 5}, {2, 10}}
Output: 2 -1

Naive Approach: The simplest approach to solve this problem is as follows:

• For a query of type 1, then update arr[X – 1] to Y.
• Otherwise, traverse the array and print the position of the first array element which is greater than or equal to K.

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

Efficient Approach: The above approach can be optimized by using segment tree. The idea is to build and update the tree using the concept of Range Maximum Query with Node Update. Follow the steps below to solve the problem:

• Build a segment tree with each node consisting of the maximum of its subtree.
• Update operation can be performed by using the concept of Range Maximum Query with Node Update
• Position of the first array element which is greater than or equal to K can be found by recursively checking for the following conditions: 
  • Check if the root of the left subtree is greater than or equal to K or not. If found to be true, then find the position from the left subtree. If no such array element is found in the left subtree, then recursively find the position in the right subtree.
  • Otherwise, recursively find the position in the right subtree.
• Finally, print the position of an array element which is greater than or equal to K.

Below is the implementation of the above approach :

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the mid
// of start and end
int getMid(int s, int e) { return s + (e - s) / 2; }

// Function to update nodes at position index
void updateValue(int arr[], int* st, int ss, int se,
                 int index, int value, int node)
{

    // If index is out of range
    if (index < ss || index > se) {
        cout << "Invalid Input" << endl;
        return;
    }

    // If a leaf node is found
    if (ss == se) {

        // update value in array
        arr[index] = value;

        // Update value in
        // the segment tree
        st[node] = value;
    }
    else {

        // Stores mid of ss and se
        int mid = getMid(ss, se);

        // If index is less than or
        // equal to mid
        if (index >= ss && index <= mid) {

            // Recursively call for left subtree
            updateValue(arr, st, ss, mid, index, value,
                        2 * node + 1);
        }
        else {

            // Recursively call for right subtree
            updateValue(arr, st, mid + 1, se, index, value,
                        2 * node + 2);
        }

        // Update st[node]
        st[node] = max(st[2 * node + 1], st[2 * node + 2]);
    }
    return;
}

// Function to find the position of first element
// which is greater than or equal to X
int findMinimumIndex(int* st, int ss, int se, int K, int si)
{

    // If no such element found in current
    // subtree which is greater than or
    // equal to K
    if (st[si] < K)
        return 1e9;

    // If current node is leaf node
    if (ss == se) {

        // If value of current node
        // is greater than or equal to X
        if (st[si] >= K) {

            return ss;
        }

        return 1e9;
    }

    // Stores mid of ss and se
    int mid = getMid(ss, se);

    int l = 1e9;

    // If root of left subtree is
    // greater than or equal to K
    if (st[2 * si + 1] >= K)
        l = min(l, findMinimumIndex(st, ss, mid, K,
                                    2 * si + 1));

    // If no such array element is
    // found in the left subtree
    if (l == 1e9 && st[2 * si + 2] >= K)
        l = min(l, findMinimumIndex(st, mid + 1, se, K,
                                    2 * si + 2));

    return l;
}

// Function to build a segment tree
int Build(int arr[], int ss, int se, int* st, int si)
{

    // If current node is leaf node
    if (ss == se) {
        st[si] = arr[ss];
        return arr[ss];
    }

    // store mid of ss and se
    int mid = getMid(ss, se);

    // Stores maximum of left subtree and rightsubtree
    st[si] = max(Build(arr, ss, mid, st, si * 2 + 1),
                 Build(arr, mid + 1, se, st, si * 2 + 2));

    return st[si];
}

// Function to initialize a segment tree
// for the given array
int* constructST(int arr[], int n)
{

    // Height of segment tree
    int x = (int)(ceil(log2(n)));

    // Maximum size of segment tree
    int max_size = 2 * (int)pow(2, x) - 1;

    // Allocate memory
    int* st = new int[max_size];

    // Fill the allocated memory st
    Build(arr, 0, n - 1, st, 0);

    // Return the constructed segment tree
    return st;
}

// Function to perform the queries of
// the given type
void PerformQueries(int arr[], int N,
                    vector<vector<int> > Q)
{

    // Build segment tree for the given array
    int* st = constructST(arr, N);

    // Traverse the query array
    for (int i = 0; i < Q.size(); i++) {

        // If query of type 1 found
        if (Q[i][0] == 1)

            updateValue(arr, st, 0, N - 1, Q[i][1] - 1, 5,
                        0);
        else {

            // Stores index of first array element
            // which is greater than or equal
            // to Q[i][1]
            int f = findMinimumIndex(st, 0, N - 1, Q[i][1],
                                     0);
            if (f < N)
                cout << f + 1 << " ";
            else
                cout << -1 << " ";
        }
    }
}

// Driver Code
int main()
{
    int arr[] = { 1, 3, 2, 4, 6 };

    vector<vector<int> > Q{
        { 2, 5 }, { 1, 3, 5 }, { 2, 4 }, { 2, 8 }
    };
    int N = sizeof(arr) / sizeof(arr[0]);

    PerformQueries(arr, N, Q);
    return 0;
}

Java

// Java program to implement
// the above approach

import java.io.*;
class GFG
{

  // Function to find the mid
  // of start and end
  static int getMid(int s, int e)
  {
    return s + (e - s) / 2;
  }
  static void updateValue(int arr[], int[] st, int ss,
                          int se, int index, int value,
                          int node)
  {

    // If index is out of range
    if (index < ss || index > se)
    {
      System.out.println("Invalid Input");
      return;
    }

    // If a leaf node is found
    if (ss == se)
    {

      // update value in array
      arr[index] = value;

      // Update value in
      // the segment tree
      st[node] = value;
    }
    else
    {

      // Stores mid of ss and se
      int mid = getMid(ss, se);

      // If index is less than or
      // equal to mid
      if (index >= ss && index <= mid)
      {

        // Recursively call for left subtree
        updateValue(arr, st, ss, mid, index, value,
                    2 * node + 1);
      }
      else
      {

        // Recursively call for right subtree
        updateValue(arr, st, mid + 1, se, index,
                    value, 2 * node + 2);
      }

      // Update st[node]
      st[node] = Math.max(st[2 * node + 1],
                          st[2 * node + 2]);
    }
  }

  // Function to find the position of first element
  // which is greater than or equal to X
  static int findMinimumIndex(int[] st, int ss, int se,
                              int K, int si)
  {

    // If no such element found in current
    // subtree which is greater than or
    // equal to K
    if (st[si] < K)
      return 1000000000;

    // If current node is leaf node
    if (ss == se)
    {

      // If value of current node
      // is greater than or equal to X
      if (st[si] >= K)
      {
        return ss;
      }
      return 1000000000;
    }

    // Stores mid of ss and se
    int mid = getMid(ss, se);

    int l = 1000000000;

    // If root of left subtree is
    // greater than or equal to K
    if (st[2 * si + 1] >= K)
      l = Math.min(l, findMinimumIndex(st, ss, mid, K,
                                       2 * si + 1));

    // If no such array element is
    // found in the left subtree
    if (l == 1e9 && st[2 * si + 2] >= K)
      l = Math.min(l,
                   findMinimumIndex(st, mid + 1, se,
                                    K, 2 * si + 2));

    return l;
  }

  // Function to build a segment tree
  static int Build(int arr[], int ss, int se, int[] st,
                   int si)
  {

    // If current node is leaf node
    if (ss == se)
    {
      st[si] = arr[ss];
      return arr[ss];
    }

    // store mid of ss and se
    int mid = getMid(ss, se);

    // Stores maximum of left subtree and rightsubtree
    st[si] = Math.max(
      Build(arr, ss, mid, st, si * 2 + 1),
      Build(arr, mid + 1, se, st, si * 2 + 2));

    return st[si];
  }

  // Function to initialize a segment tree
  // for the given array
  static int[] constructST(int arr[], int n)
  {

    // Height of segment tree
    int x = (int)Math.ceil(Math.log(n) / Math.log(2));

    // Maximum size of segment tree
    int max_size = 2 * (int)Math.pow(2, x) - 1;

    // Allocate memory
    int[] st = new int[max_size];

    // Fill the allocated memory st
    Build(arr, 0, n - 1, st, 0);

    // Return the constructed segment tree
    return st;
  }
  static void PerformQueries(int arr[], int N, int[][] Q)
  {

    // Build segment tree for the given array
    int[] st = constructST(arr, N);

    // Traverse the query array
    for (int i = 0; i < Q.length; i++)
    {

      // If query of type 1 found
      if (Q[i][0] == 1)
        updateValue(arr, st, 0, N - 1, Q[i][1] - 1,
                    5, 0);
      else {

        // Stores index of first array element
        // which is greater than or equal
        // to Q[i][1]
        int f = findMinimumIndex(st, 0, N - 1,
                                 Q[i][1], 0);
        if (f < N)
          System.out.print(f + 1 + " ");
        else
          System.out.print(-1 + " ");
      }
    }
  }

  // Driver Code
  public static void main(String[] args)
  {
    int arr[] = { 1, 3, 2, 4, 6 };

    int[][] Q
      = { { 2, 5 }, { 1, 3, 5 }, { 2, 4 }, { 2, 8 } };
    int N = arr.length;
    PerformQueries(arr, N, Q);
  }
}

// This code is contributed by hemanthsawarna1506

Python3

# Python3 program to implement
# the above approach
import math

# Function to find the mid
# of start and end
def getMid(s,e):
    return s + math.floor((e - s) / 2)

def updateValue(arr,st,ss,se,index,value,node):
    # If index is out of range
    if (index < ss or index > se):
      print("Invalid Input")
      return

    # If a leaf node is found
    if (ss == se):
      # update value in array
      arr[index] = value

      # Update value in
      # the segment tree
      st[node] = value
    else:
      # Stores mid of ss and se
      mid = getMid(ss, se)

      # If index is less than or
      # equal to mid
      if (index >= ss and index <= mid):

        # Recursively call for left subtree
        updateValue(arr, st, ss, mid, index, value, 2 * node + 1)
      else:

        # Recursively call for right subtree
        updateValue(arr, st, mid + 1, se, index, value, 2 * node + 2)

      # Update st[node]
      st[node] = max(st[2 * node + 1], st[2 * node + 2])

# Function to find the position of first element
# which is greater than or equal to X
def findMinimumIndex(st, ss, se, K, si):
    # If no such element found in current
    # subtree which is greater than or
    # equal to K
    if (st[si] < K):
      return 1000000000

    # If current node is leaf node
    if (ss == se):
      # If value of current node
      # is greater than or equal to X
      if (st[si] >= K):
        return ss
      return 1000000000

    # Stores mid of ss and se
    mid = getMid(ss, se)

    l = 1000000000

    # If root of left subtree is
    # greater than or equal to K
    if (st[2 * si + 1] >= K):
      l = min(l, findMinimumIndex(st, ss, mid, K, 2 * si + 1))

    # If no such array element is
    # found in the left subtree
    if (l == 1e9 and st[2 * si + 2] >= K):
      l = min(l, findMinimumIndex(st, mid + 1, se, K, 2 * si + 2))

    return l

# Function to build a segment tree
def Build(arr, ss, se, st, si):

    # If current node is leaf node
    if (ss == se):
      st[si] = arr[ss]
      return arr[ss]

    # store mid of ss and se
    mid = getMid(ss, se)

    # Stores maximum of left subtree and rightsubtree
    st[si] = max(Build(arr, ss, mid, st, si * 2 + 1), Build(arr, mid + 1, se, st, si * 2 + 2))

    return st[si]

# Function to initialize a segment tree
# for the given array
def constructST(arr,n):

    # Height of segment tree
    x = math.ceil(math.log(n) / math.log(2))

    # Maximum size of segment tree
    max_size = 2 * pow(2, x) - 1

    # Allocate memory
    st = [0]*(max_size)

    # Fill the allocated memory st
    Build(arr, 0, n - 1, st, 0)

    # Return the constructed segment tree
    return st

def PerformQueries(arr, N, Q):

    # Build segment tree for the given array
    st = constructST(arr, N)

    # Traverse the query array
    for i in range(len(Q)):

      # If query of type 1 found
      if (Q[i][0] == 1):
        updateValue(arr, st, 0, N - 1, Q[i][1] - 1, 5, 0)
      else:
        # Stores index of first array element
        # which is greater than or equal
        # to Q[i][1]
        f = findMinimumIndex(st, 0, N - 1, Q[i][1], 0)
        if (f < N):
          print((f + 1 ), end = " ")
        else:
          print(-1, end = " ")

# Driver code
arr = [1, 3, 2, 4, 6 ]
Q= [[ 2, 5 ], [ 1, 3, 5 ], [2, 4 ], [ 2, 8 ]]
N = len(arr)

PerformQueries(arr, N, Q)

# This code is contributed by decode2207.

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function to find the mid
// of start and end
static int getMid(int s, int e)
{
    return s + (e - s) / 2;
}

static void updateValue(int[] arr, int[] st, int ss,
                        int se, int index, int value,
                        int node)
{

    // If index is out of range
    if (index < ss || index > se)
    {
        Console.WriteLine("Invalid Input");
        return;
    }

    // If a leaf node is found
    if (ss == se)
    {

        // Update value in array
        arr[index] = value;

        // Update value in
        // the segment tree
        st[node] = value;
    }
    else
    {

        // Stores mid of ss and se
        int mid = getMid(ss, se);

        // If index is less than or
        // equal to mid
        if (index >= ss && index <= mid)
        {

            // Recursively call for left subtree
            updateValue(arr, st, ss, mid, index, value,
                        2 * node + 1);
        }
        else
        {

            // Recursively call for right subtree
            updateValue(arr, st, mid + 1, se, index,
                        value, 2 * node + 2);
        }

        // Update st[node]
        st[node] = Math.Max(st[2 * node + 1],
                            st[2 * node + 2]);
    }
}

// Function to find the position of first element
// which is greater than or equal to X
static int findMinimumIndex(int[] st, int ss, int se,
                            int K, int si)
{

    // If no such element found in current
    // subtree which is greater than or
    // equal to K
    if (st[si] < K)
        return 1000000000;

    // If current node is leaf node
    if (ss == se)
    {

        // If value of current node
        // is greater than or equal to X
        if (st[si] >= K)
        {
            return ss;
        }
        return 1000000000;
    }

    // Stores mid of ss and se
    int mid = getMid(ss, se);

    int l = 1000000000;

    // If root of left subtree is
    // greater than or equal to K
    if (st[2 * si + 1] >= K)
        l = Math.Min(l, findMinimumIndex(st, ss, mid, K,
                                         2 * si + 1));

    // If no such array element is
    // found in the left subtree
    if (l == 1e9 && st[2 * si + 2] >= K)
        l = Math.Min(l,
                     findMinimumIndex(st, mid + 1, se,
                                    K, 2 * si + 2));

    return l;
}

// Function to build a segment tree
static int Build(int[] arr, int ss, int se,
                 int[] st, int si)
{

    // If current node is leaf node
    if (ss == se)
    {
        st[si] = arr[ss];
        return arr[ss];
    }

    // Store mid of ss and se
    int mid = getMid(ss, se);

    // Stores maximum of left subtree and rightsubtree
    st[si] = Math.Max(
        Build(arr, ss, mid, st, si * 2 + 1),
        Build(arr, mid + 1, se, st, si * 2 + 2));

    return st[si];
}

// Function to initialize a segment tree
// for the given array
static int[] constructST(int[] arr, int n)
{

    // Height of segment tree
    int x = (int)Math.Ceiling(Math.Log(n) /
                              Math.Log(2));

    // Maximum size of segment tree
    int max_size = 2 * (int)Math.Pow(2, x) - 1;

    // Allocate memory
    int[] st = new int[max_size];

    // Fill the allocated memory st
    Build(arr, 0, n - 1, st, 0);

    // Return the constructed segment tree
    return st;
}

static void PerformQueries(int[] arr, int N, int[][] Q)
{

    // Build segment tree for the given array
    int[] st = constructST(arr, N);

    // Traverse the query array
    for(int i = 0; i < Q.Length; i++)
    {

        // If query of type 1 found
        if (Q[i][0] == 1)
            updateValue(arr, st, 0, N - 1,
                        Q[i][1] - 1, 5, 0);
        else
        {

            // Stores index of first array element
            // which is greater than or equal
            // to Q[i][1]
            int f = findMinimumIndex(st, 0, N - 1,
                                     Q[i][1], 0);
            if (f < N)
                Console.Write(f + 1 + " ");
            else
                Console.Write(-1 + " ");
        }
    }
}

// Driver Code
public static void Main(string[] args)
{
    int[] arr = { 1, 3, 2, 4, 6 };
    int[][] Q = new int[4][];

    // Initialize the elements
    Q[0] = new int[] { 2, 5 };
    Q[1] = new int[] { 1, 3, 5 };
    Q[2] = new int[] { 2, 4 };
    Q[3] = new int[] { 2, 8 };

    int N = arr.Length;
    PerformQueries(arr, N, Q);
}
}

// This code is contributed by ukasp

Javascript

<script>
// Javascript program to implement
// the above approach

// Function to find the mid
// of start and end
function getMid(s,e)
{
    return s + Math.floor((e - s) / 2);
}

function updateValue(arr,st,ss,se,index,value,node)
{
    // If index is out of range
    if (index < ss || index > se)
    {
      document.write("Invalid Input<br>");
      return;
    }

    // If a leaf node is found
    if (ss == se)
    {

      // update value in array
      arr[index] = value;

      // Update value in
      // the segment tree
      st[node] = value;
    }
    else
    {

      // Stores mid of ss and se
      let mid = getMid(ss, se);

      // If index is less than or
      // equal to mid
      if (index >= ss && index <= mid)
      {

        // Recursively call for left subtree
        updateValue(arr, st, ss, mid, index, value,
                    2 * node + 1);
      }
      else
      {

        // Recursively call for right subtree
        updateValue(arr, st, mid + 1, se, index,
                    value, 2 * node + 2);
      }

      // Update st[node]
      st[node] = Math.max(st[2 * node + 1],
                          st[2 * node + 2]);
    }
}

// Function to find the position of first element
// which is greater than or equal to X
function findMinimumIndex(st,ss,se,K,si)
{
    // If no such element found in current
    // subtree which is greater than or
    // equal to K
    if (st[si] < K)
      return 1000000000;

    // If current node is leaf node
    if (ss == se)
    {

      // If value of current node
      // is greater than or equal to X
      if (st[si] >= K)
      {
        return ss;
      }
      return 1000000000;
    }

    // Stores mid of ss and se
    let mid = getMid(ss, se);

    let l = 1000000000;

    // If root of left subtree is
    // greater than or equal to K
    if (st[2 * si + 1] >= K)
      l = Math.min(l, findMinimumIndex(st, ss, mid, K,
                                       2 * si + 1));

    // If no such array element is
    // found in the left subtree
    if (l == 1e9 && st[2 * si + 2] >= K)
      l = Math.min(l,
                   findMinimumIndex(st, mid + 1, se,
                                    K, 2 * si + 2));

    return l;
}

// Function to build a segment tree
function Build(arr,ss,se,st,si)
{
    // If current node is leaf node
    if (ss == se)
    {
      st[si] = arr[ss];
      return arr[ss];
    }

    // store mid of ss and se
    let mid = getMid(ss, se);

    // Stores maximum of left subtree and rightsubtree
    st[si] = Math.max(
      Build(arr, ss, mid, st, si * 2 + 1),
      Build(arr, mid + 1, se, st, si * 2 + 2));

    return st[si];
}

// Function to initialize a segment tree
  // for the given array
function constructST(arr,n)
{
    // Height of segment tree
    let x = Math.ceil(Math.log(n) / Math.log(2));

    // Maximum size of segment tree
    let max_size = 2 * Math.pow(2, x) - 1;

    // Allocate memory
    let st = new Array(max_size);

    // Fill the allocated memory st
    Build(arr, 0, n - 1, st, 0);

    // Return the constructed segment tree
    return st;
}

function PerformQueries(arr,N,Q)
{
    // Build segment tree for the given array
    let st = constructST(arr, N);

    // Traverse the query array
    for (let i = 0; i < Q.length; i++)
    {

      // If query of type 1 found
      if (Q[i][0] == 1)
        updateValue(arr, st, 0, N - 1, Q[i][1] - 1,
                    5, 0);
      else {

        // Stores index of first array element
        // which is greater than or equal
        // to Q[i][1]
        let f = findMinimumIndex(st, 0, N - 1,
                                 Q[i][1], 0);
        if (f < N)
          document.write((f + 1 )+ " ");
        else
          document.write(-1 + " ");
      }
    }
}

// Driver Code
let arr=[1, 3, 2, 4, 6 ];
let Q= [[ 2, 5 ], [ 1, 3, 5 ], [2, 4 ], [ 2, 8 ]];
let N = arr.length;
PerformQueries(arr, N, Q);


// This code is contributed by avanitrachhadiya2155
</script>

Output:

5 3 -1 

Time Complexity: O(n*log(n) + q*log(n))
Auxiliary Space: O(n*log(n))