Given an integer array arr[], and some queries consisting of an integer K, the task is to determine if its possible to make all the integers positive by flipping signs of integers exactly K times. We can flip the sign of an integer more than once. If possible, then print Yes else print No.
Examples:
Input: arr[] = {-1, 2, -3, 4, 5}, q[] = {1, 2}
Output:
No
Yes
Query 1: Only the sign of either -1 or -3
can be changed and not both.
Query 2: Signs of both the negative numbers
can be changed to positive.
Input: arr[] = {-1, -1, 0, 6}, q[] = {1, 2, 3, 4}
Output:
No
Yes
Yes
Yes
Approach: Following will be the algorithm that we will use:
- Count number of negative integers in the array and store it in a variable cnt.
- If there is no zero in the array:
- If K ? cnt then answer will be Yes.
- If K = cnt and (K – cnt) % 2 = 0 then answer will be Yes.
- Else answer will be No.
- If there exists a zero in the array then the answer will only be No if K < cnt else the answer is always Yes.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// To store the count of// negative integersint cnt_neg;// Check if zero existsbool exists_zero;// Function to find the count of// negative integers and check// if zero exists in the arrayvoid preProcess(int arr[], int n){ for (int i = 0; i < n; i++) { if (arr[i] < 0) cnt_neg++; if (arr[i] == 0) exists_zero = true; }}// Function that returns true if array// elements can be made positive by// changing signs exactly k timesbool isPossible(int k){ if (!exists_zero) { if (k >= cnt_neg and (k - cnt_neg) % 2 == 0) return true; else return false; } else { if (k >= cnt_neg) return true; else return false; }}// Driver codeint main(){ int arr[] = { -1, 2, -3, 4, 5 }; int n = sizeof(arr) / sizeof(int); // Pre-processing preProcess(arr, n); int queries[] = { 1, 2, 3, 4 }; int q = sizeof(queries) / sizeof(int); // Perform queries for (int i = 0; i < q; i++) { if (isPossible(queries[i])) cout << "Yes" << endl; else cout << "No" << endl; } return 0;} |
Java
// Java implementation of the approachimport java.io.*;class GFG {// To store the count of// negative integersstatic int cnt_neg;// Check if zero existsstatic boolean exists_zero;// Function to find the count of// negative integers and check// if zero exists in the arraystatic void preProcess(int []arr, int n){ for (int i = 0; i < n; i++) { if (arr[i] < 0) cnt_neg++; if (arr[i] == 0) exists_zero = true; }}// Function that returns true if array// elements can be made positive by// changing signs exactly k timesstatic boolean isPossible(int k){ if (!exists_zero) { if (k >= cnt_neg && (k - cnt_neg) % 2 == 0) return true; else return false; } else { if (k >= cnt_neg) return true; else return false; }}// Driver codepublic static void main (String[] args) { int arr[] = { -1, 2, -3, 4, 5 }; int n = arr.length; // Pre-processing preProcess(arr, n); int queries[] = { 1, 2, 3, 4 }; int q = arr.length; // Perform queries for (int i = 0; i < q; i++) { if (isPossible(queries[i])) System.out.println( "Yes"); else System.out.println( "No"); }}}// This code is contributed by anuj_67.. |
Python3
# Python3 implementation of the approach # To store the count of # negative integers cnt_neg = 0; # Check if zero exists exists_zero = None; # Function to find the count of # negative integers and check # if zero exists in the array def preProcess(arr, n) : global cnt_neg for i in range(n) : if (arr[i] < 0) : cnt_neg += 1; if (arr[i] == 0) : exists_zero = True; # Function that returns true if array # elements can be made positive by # changing signs exactly k times def isPossible(k) : if (not exists_zero) : if (k >= cnt_neg and (k - cnt_neg) % 2 == 0) : return True; else : return False; else : if (k >= cnt_neg) : return True; else : return False; # Driver code if __name__ == "__main__" : arr = [ -1, 2, -3, 4, 5 ]; n = len(arr); # Pre-processing preProcess(arr, n); queries = [ 1, 2, 3, 4 ]; q = len(queries); # Perform queries for i in range(q) : if (isPossible(queries[i])) : print("Yes"); else : print("No"); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System;class GFG{ // To store the count of// negative integersstatic int cnt_neg;// Check if zero existsstatic bool exists_zero ;// Function to find the count of// negative integers and check// if zero exists in the arraystatic void preProcess(int []arr, int n){ for (int i = 0; i < n; i++) { if (arr[i] < 0) cnt_neg++; if (arr[i] == 0) exists_zero = true; }}// Function that returns true if array// elements can be made positive by// changing signs exactly k timesstatic bool isPossible(int k){ if (!exists_zero) { if (k >= cnt_neg && (k - cnt_neg) % 2 == 0) return true; else return false; } else { if (k >= cnt_neg) return true; else return false; }}// Driver codestatic public void Main (){ int []arr = { -1, 2, -3, 4, 5 }; int n = arr.Length; // Pre-processing preProcess(arr, n); int []queries = { 1, 2, 3, 4 }; int q = arr.Length; // Perform queries for (int i = 0; i < q; i++) { if (isPossible(queries[i])) Console.WriteLine( "Yes"); else Console.WriteLine( "No"); }}}// This code is contributed by ajit... |
Javascript
<script>// Javascript implementation of the approach// To store the count of// negative integerslet cnt_neg = 0;// Check if zero existslet exists_zero = false;// Function to find the count of// negative integers and check// if zero exists in the arrayfunction preProcess(arr, n){ for (let i = 0; i < n; i++) { if (arr[i] < 0) cnt_neg++; if (arr[i] == 0) exists_zero = true; }}// Function that returns true if array// elements can be made positive by// changing signs exactly k timesfunction isPossible(k){ if (!exists_zero) { if (k >= cnt_neg && (k - cnt_neg) % 2 == 0) return true; else return false; } else { if (k >= cnt_neg) return true; else return false; }}// Driver code let arr = [ -1, 2, -3, 4, 5 ]; let n = arr.length; // Pre-processing preProcess(arr, n); let queries = [ 1, 2, 3, 4 ]; let q = queries.length; // Perform queries for (let i = 0; i < q; i++) { if (isPossible(queries[i])) document.write("Yes<br>"); else document.write("No<br>"); }</script> |
PHP
<?php// To store the count of// negative integers$cnt_neg = 0;// Check if zero exists$exists_zero = false;// Function to find the count of// negative integers and check// if zero exists in the arrayfunction preProcess($arr, $n) { global $cnt_neg, $exists_zero; for ($i = 0; $i < $n; $i++) { if ($arr[$i] < 0) { $cnt_neg++; } if ($arr[$i] == 0) { $exists_zero = true; } }}// Function that returns true if array// elements can be made positive by// changing signs exactly k timesfunction isPossible($k) { global $cnt_neg, $exists_zero; if (!$exists_zero) { if ($k >= $cnt_neg && ($k - $cnt_neg) % 2 == 0) { return true; } else { return false; } } else { if ($k >= $cnt_neg) { return true; } else { return false; } }}// Driver code$arr = array(-1, 2, -3, 4, 5);$n = sizeof($arr) / sizeof($arr[0]);// Pre-processingpreProcess($arr, $n);$queries = array(1, 2, 3, 4);$q = sizeof($queries) / sizeof($queries[0]);// Perform queriesfor ($i = 0; $i < $q; $i++) { if (isPossible($queries[$i])) { echo "Yes\n"; } else { echo "No\n"; }}?> |
Output:
No Yes No Yes
Time Complexity: O(n + q), where n and q represents the size of the given arrays.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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