Given a variable X having an initial value 0 and an array of queries Q[] of size N containing the type of operations, the task is to return the value of N after performing all the below operations:
- Type-1: Increment the value of X by 1.
- Type-2: Decrement the value of X by 1.
Examples:
Input: Q = {2, 1, 1}
Output: 1
Explanation: The operations are performed as follow:
Initially, X = 0,
Query 1(Type 2): X is decremented by 1, X = 0 – 1 = -1
Query 2(Type 1): X is incremented by 1, X = -1 + 1 = 0
Query 3(Type 1): X is incremented by 1, X = 0 + 1 = 1
Hence, the output will be 1Input: Q = {1, 1, 1}
Output: 3
Approach: To solve the problem follow the below idea:
- Traverse the given array of Queries.
- Whenever the array element is 1, increment N by 1, and
- Whenever the array element is 2, decrement N by 1.
- Return the final value of N as the required answer.
Below is the implementation for the above approach:
C++
// C++ code for the above approach.#include <bits/stdc++.h>using namespace std;// Function to findvalue after queriesint findvalue(vector<int>& queries){ // Storing array size in n int n = queries.size(); int x = 0; for (int i = 0; i < n; i++) { // If query is of type 1, increment if (queries[i] == 1) { x = x + 1; } // Else query is of type 2, decrement else { x = x - 1; } } return x;}// Drivers codeint main(){ // Declaring array of integers // which contains queries vector<int> queries = { 2, 1, 1 }; int ans = findvalue(queries); // Function Call cout << ans; return 0;} |
Java
// Java code for the above approach.public class GFG { // Function to findvalue after queries static int findvalue(int []queries) { // Storing array size in n int n = queries.length; int x = 0; for (int i = 0; i < n; i++) { // If query is of type 1, increment if (queries[i] == 1) { x = x + 1; } // Else query is of type 2, decrement else { x = x - 1; } } return x; } // Drivers code public static void main(String[] args) { // Declaring array of integers // which contains queries int []queries = { 2, 1, 1 }; int ans = findvalue(queries); // Function Call System.out.println(ans); }}// This code is contributed by AnkThon |
Python3
# Python3 code for the above approach.# Function to findvalue after queriesdef findvalue(queries) : # Storing array size in n n = len(queries); x = 0; for i in range(n) : # If query is of type 1, increment if (queries[i] == 1) : x = x + 1; # Else query is of type 2, decrement else : x = x - 1; return x;# Drivers codeif __name__ == "__main__" : # Declaring array of integers # which contains queries queries = [ 2, 1, 1 ]; ans = findvalue(queries); # Function Call print(ans); # This code is contributed by AnkThon |
C#
// C# code for the above approach.using System;using System.Collections.Generic;class GFG { // Function to findvalue after queries static int findvalue(List<int> queries) { // Storing array size in n int n = queries.Count; int x = 0; for (int i = 0; i < n; i++) { // If query is of type 1, increment if (queries[i] == 1) { x = x + 1; } // Else query is of type 2, decrement else { x = x - 1; } } return x; } // Drivers code public static void Main(string[] args) { // Declaring array of integers // which contains queries List<int> queries = new List<int>{ 2, 1, 1 }; int ans = findvalue(queries); // Function Call Console.Write(ans); }}// This code is contributed by phasing17 |
Javascript
<script>// Javascript code for the above approach.// Function to findvalue after queriesfunction findvalue( queries){ // Storing array size in n let n = queries.length; let x = 0; for (let i = 0; i < n; i++) { // If query is of type 1, increment if (queries[i] == 1) { x = x + 1; } // Else query is of type 2, decrement else { x = x - 1; } } return x;}// Drivers code // Declaring array of integers // which contains queries let queries = [ 2, 1, 1 ]; let ans = findvalue(queries); // Function Call document.write(ans); // This code is contributed by satwik4409. </script> |
Output
1
Time Complexity: O(N).
Auxiliary Space: O(1).
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