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 < 2Explanation:

The inequality formed is consistent and valid.

Input:Integers: [ 8, 34, 25, 1, -5, 10], Signs: [ >, >, <, <, > ]Output:34 > 25 > -5 < 1 < 10 > 8Explanation:

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:

## 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); ` ` ` `} ` `} ` |

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## 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 ` |

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**Output:**

0 < 5 > 1 < 2

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

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