Given an integer array *H* which represents heights of buildings and an integer *K*. The task is to reach the last building from the first with the following rules:

- Getting to a building of height
**H**from a building of height_{j}**H**is only possible if_{i}**|H**and the building appears one after another in the array._{i}– H_{j}| <= K - If getting to a building is not possible then a number of buildings of intermediate heights can be inserted in between the two building.

Find the minimum number of insertions done in step 2 to make it possible to go from first building to the last.**Examples:**

Input:H[] = {2, 4, 8, 16}, K = 3Output:3

Add 1 building of height 5 between buildings of height 4 and 8.

And add 2 buildings of heights 11 and 14 respectively between buildings of height 8 and 16.Input:H[] = {5, 55, 100, 1000}, K = 10Output:97

**Approach:**

- Run a loop from
*1 to n-1*and check whether abs(H[i] – H[i-1]) <= K. - If the above condition is true then skip to the next iteration of the loop.
- If the condition is false, then the insertions required will be equal to ceil(diff / K) – 1 where diff = abs(H[i] – H[i-1])
- Print the total insertions in the end.

Below is the implementation of above approach:

## C++

`// CPP implementation of above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to return minimum` `// number of insertions required` `int` `minInsertions(` `int` `H[], ` `int` `n, ` `int` `K)` `{` ` ` `// Initialize insertions to 0` ` ` `int` `inser = 0;` ` ` `for` `(` `int` `i = 1; i < n; ++i) {` ` ` `float` `diff = ` `abs` `(H[i] - H[i - 1]);` ` ` `if` `(diff <= K)` ` ` `continue` `;` ` ` `else` ` ` `inser += ` `ceil` `(diff / K) - 1;` ` ` `}` ` ` `// return total insertions` ` ` `return` `inser;` `}` `// Driver program` `int` `main()` `{` ` ` `int` `H[] = { 2, 4, 8, 16 }, K = 3;` ` ` `int` `n = ` `sizeof` `(H) / ` `sizeof` `(H[0]);` ` ` `cout << minInsertions(H, n, K);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of above approach` `class` `GFG{` `// Function to return minimum` `// number of insertions required` `static` `int` `minInsertions(` `int` `[] H, ` `int` `n, ` `int` `K)` `{` ` ` `// Initialize insertions to 0` ` ` `int` `inser = ` `0` `;` ` ` `for` `(` `int` `i = ` `1` `; i < n; ++i) {` ` ` `float` `diff = Math.abs(H[i] - H[i - ` `1` `]);` ` ` `if` `(diff <= K)` ` ` `continue` `;` ` ` `else` ` ` `inser += Math.ceil(diff / K) - ` `1` `;` ` ` `}` ` ` `// return total insertions` ` ` `return` `inser;` `}` `// Driver program` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `[] H = ` `new` `int` `[]{ ` `2` `, ` `4` `, ` `8` `, ` `16` `};` ` ` `int` `K = ` `3` `;` ` ` `int` `n = H.length;` ` ` `System.out.println(minInsertions(H, n, K));` `}` `}` `// This code is contributed by mits` |

## Python3

`# Python3 implementation of above approach` `import` `math` `# Function to return minimum` `# number of insertions required` `def` `minInsertions(H, n, K):` ` ` `# Initialize insertions to 0` ` ` `inser ` `=` `0` `;` ` ` `for` `i ` `in` `range` `(` `1` `, n):` ` ` `diff ` `=` `abs` `(H[i] ` `-` `H[i ` `-` `1` `]);` ` ` `if` `(diff <` `=` `K):` ` ` `continue` `;` ` ` `else` `:` ` ` `inser ` `+` `=` `math.ceil(diff ` `/` `K) ` `-` `1` `;` ` ` `# return total insertions` ` ` `return` `inser;` `# Driver Code` `H ` `=` `[` `2` `, ` `4` `, ` `8` `, ` `16` `];` `K ` `=` `3` `;` `n ` `=` `len` `(H);` `print` `(minInsertions(H, n, K));` `# This code is contributed` `# by mits` |

## C#

`// C# implementation of above approach` `using` `System;` `class` `GFG` `{` `// Function to return minimum` `// number of insertions required` `static` `int` `minInsertions(` `int` `[] H,` ` ` `int` `n, ` `int` `K)` `{` ` ` `// Initialize insertions to 0` ` ` `int` `inser = 0;` ` ` `for` `(` `int` `i = 1; i < n; ++i)` ` ` `{` ` ` `float` `diff = Math.Abs(H[i] - H[i - 1]);` ` ` `if` `(diff <= K)` ` ` `continue` `;` ` ` `else` ` ` `inser += (` `int` `)Math.Ceiling(diff / K) - 1;` ` ` `}` ` ` `// return total insertions` ` ` `return` `inser;` `}` `// Driver Code` `static` `void` `Main()` `{` ` ` `int` `[] H = ` `new` `int` `[]{ 2, 4, 8, 16 };` ` ` `int` `K = 3;` ` ` `int` `n = H.Length;` ` ` `Console.WriteLine(minInsertions(H, n, K));` `}` `}` `// This code is contributed by mits` |

## PHP

`<?php` `// PHP implementation of above approach` `// Function to return minimum` `// number of insertions required` `function` `minInsertions(` `$H` `, ` `$n` `, ` `$K` `)` `{` ` ` `// Initialize insertions to 0` ` ` `$inser` `= 0;` ` ` `for` `(` `$i` `= 1; ` `$i` `< ` `$n` `; ++` `$i` `)` ` ` `{` ` ` `$diff` `= ` `abs` `(` `$H` `[` `$i` `] - ` `$H` `[` `$i` `- 1]);` ` ` `if` `(` `$diff` `<= ` `$K` `)` ` ` `continue` `;` ` ` `else` ` ` `$inser` `+= ` `ceil` `(` `$diff` `/ ` `$K` `) - 1;` ` ` `}` ` ` `// return total insertions` ` ` `return` `$inser` `;` `}` `// Driver Code` `$H` `= ` `array` `(2, 4, 8, 16 );` `$K` `= 3;` `$n` `= sizeof(` `$H` `);` `echo` `minInsertions(` `$H` `, ` `$n` `, ` `$K` `);` `// This code is contributed` `// by Akanksha Rai(Abby_akku)` `?>` |

## Javascript

`<script>` `// Javascript implementation of above approach` `// Function to return minimum` `// number of insertions required` `function` `minInsertions( H, n, K)` `{` ` ` `// Initialize insertions to 0` ` ` `var` `inser = 0;` ` ` `for` `(` `var` `i = 1; i < n; ++i) {` ` ` `var` `diff = Math.abs(H[i] - H[i - 1]);` ` ` `if` `(diff <= K)` ` ` `continue` `;` ` ` `else` ` ` `inser += Math.ceil(diff / K) - 1;` ` ` `}` ` ` `// return total insertions` ` ` `return` `inser;` `}` `var` `H = [ 2, 4, 8, 16 ];` `var` `K = 3;` `var` `n = H.length;` `document.write(minInsertions(H, n, K));` `// This code is contributed by SoumikMondal` `</script>` |

**Output:**

3

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