# Arrange the numbers in the Array as per given inequalities

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:

1. Sort the list of integers in ascending order.
2. Maintain two variables, say low and high, pointing at the first and last index of the list of integers.
3. 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.
4. Finally, add the remaining element to the last position.

Below is the implementation of the above approach:

## Java

 `// Java program for the above approach ` `import` `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); ` `    ``} ` `} `

## 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 `

Output:

```0 < 5 > 1 < 2
```

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

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