Given a list of N distinct integers and a list of N-1 inequality signs, the task is to insert the integers between the inequality signs, such that the final inequality formed always holds true.
Note: The order of the inequality signs should not be changed.
Examples:
Input: Integers: [ 2, 5, 1, 0 ], Signs: [ <, >, < ]
Output: 0 < 5 > 1 < 2
Explanation:
The inequality formed is consistent and valid.Input: Integers: [ 8, 34, 25, 1, -5, 10], Signs: [ >, >, <, <, > ]
Output: 34 > 25 > -5 < 1 < 10 > 8
Explanation:
The inequality formed is consistent and valid.
Approach: The list of inequality symbols can contain symbols in any order. So to obtain a consistent inequality, put the smallest integer left in the array before each < symbol and the largest integer left before each > symbol. Based on this idea, below are the steps:
- Sort the list of integers in ascending order.
- Maintain two variables, say low and high, pointing at the first and last index of the list of integers.
- Iterate over the list of inequality symbols. If the current symbol is less, then add the integer pointed by low before <, and increment low to point to the next index. If the current symbol is greater, then add the integer pointed by high before >, and decrement high to point to the previous index.
- Finally, add the remaining element to the last position.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to place the integers// in between the inequality signsstring formAnInequality(int integers[], char inequalities[], int n1, int n2){ // Sort the integers array and // set the index of smallest // and largest element sort(integers, integers + n1); int lowerIndex = 0; int higherIndex = n1 - 1; string sb = ""; // Iterate over the inequalities for (int i = 0; i < n2; i++) { char ch = inequalities[i]; // Append the necessary // integers per symbol if (ch == '<') { sb += " "; sb += to_string(integers[lowerIndex++]); sb += " "; sb += ch; } else { sb += " "; sb += to_string(integers[higherIndex--]); sb += " "; sb += ch; } } // Add the final integer sb += " " + to_string(integers[lowerIndex]); // Return the answer return sb;}// Driver Codeint main(){ // Given List of Integers int integers[] = { 2, 5, 1, 0 }; // Given list of inequalities char inequalities[] = { '<', '>', '<' }; int n1 = 4; int n2 = 3; // Function Call string output = formAnInequality(integers, inequalities, n1, n2); // Print the output cout << output << endl;}// This code is contributed by phasing17 |
Java
// Java program for the above approachimport java.util.Arrays;public class PlacingNumbers { // Function to place the integers // in between the inequality signs static String formAnInequality(int[] integers, char[] inequalities) { // Sort the integers array and // set the index of smallest // and largest element Arrays.sort(integers); int lowerIndex = 0; int higherIndex = integers.length - 1; StringBuilder sb = new StringBuilder(); // Iterate over the inequalities for (char ch : inequalities) { // Append the necessary // integers per symbol if (ch == '<') { sb.append(" " + integers[lowerIndex++] + " " + ch); } else { sb.append(" " + integers[higherIndex--] + " " + ch); } } // Add the final integer sb.append(" " + integers[lowerIndex]); // Return the answer return sb.toString(); } // Driver Code public static void main(String[] args) { // Given List of Integers int[] integers = { 2, 5, 1, 0 }; // Given list of inequalities char[] inequalities = { '<', '>', '<' }; // Function Call String output = formAnInequality(integers, inequalities); // Print the output System.out.println(output); }} |
Python3
# Python3 program for # the above approach# Function to place the integers# in between the inequality signsdef formAnInequality(integers, inequalities): # Sort the integers array and # set the index of smallest # and largest element integers.sort() lowerIndex = 0 higherIndex = len(integers) - 1 sb = "" # Iterate over the inequalities for ch in inequalities: # Append the necessary # integers per symbol if (ch == '<'): sb += (" " + chr(integers[lowerIndex]) + " " + ch) lowerIndex += 1 else: sb += (" " + chr(integers[higherIndex]) + " " + ch) higherIndex -= 1 # Add the final integer sb += (" " + chr(integers[lowerIndex])) # Return the answer return sb# Driver Codeif __name__ == "__main__": # Given List of Integers integers = [2, 5, 1, 0] # Given list of inequalities inequalities = ['<', '>', '<'] # Function Call output = formAnInequality(integers, inequalities) # Print the output print(output)# This code is contributed by Chitranayal |
C#
// C# program for the above approach using System;using System.Text;class GFG{ // Function to place the integers // in between the inequality signs static String formAnInequality(int[] integers, char[] inequalities) { // Sort the integers array and // set the index of smallest // and largest element Array.Sort(integers); int lowerIndex = 0; int higherIndex = integers.Length - 1; StringBuilder sb = new StringBuilder(); // Iterate over the inequalities foreach(char ch in inequalities) { // Append the necessary // integers per symbol if (ch == '<') { sb.Append(" " + integers[lowerIndex++] + " " + ch); } else { sb.Append(" " + integers[higherIndex--] + " " + ch); } } // Add the readonly integer sb.Append(" " + integers[lowerIndex]); // Return the answer return sb.ToString(); } // Driver Code public static void Main(String[] args) { // Given List of ints int[] integers = { 2, 5, 1, 0 }; // Given list of inequalities char[] inequalities = { '<', '>', '<' }; // Function call String output = formAnInequality(integers, inequalities); // Print the output Console.WriteLine(output); } } // This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program for the above approach// Function to place the integers// in between the inequality signsfunction formAnInequality(integers, inequalities){ // Sort the integers array and // set the index of smallest // and largest element (integers).sort(function(a, b){return a - b;}); let lowerIndex = 0; let higherIndex = integers.length - 1; let sb = ""; // Iterate over the inequalities for(let ch = 0; ch < inequalities.length; ch++) { // Append the necessary // integers per symbol if (inequalities[ch] == '<') { sb += " " + (integers[lowerIndex++]) + " " + inequalities[ch]; } else { sb += " " + (integers[higherIndex--]) + " " + inequalities[ch]; } } // Add the final integer sb += " " + (integers[lowerIndex]); // Return the answer return sb;}// Driver Code// Given List of Integerslet integers = [ 2, 5, 1, 0 ];// Given list of inequalitieslet inequalities = [ '<', '>', '<' ];// Function Calllet output = formAnInequality( integers, inequalities); // Print the outputdocument.write(output);// This code is contributed by avanitrachhadiya2155</script> |
Output:
0 < 5 > 1 < 2
Time Complexity: O(N*log N)
Auxiliary Space: O(1)
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