# Find the number of boxes to be removed

Given an array arr[] representing a sequence of piles of boxes where each and every box has the same height of 1 unit. Given that you are on the top of the first pile and need to reach the ground by moving from each pile starting from leftmost to rightmost.

**Constraints**:

- One can move from the current pile of box to the next one when the height of the next pile is equal or less than the height of the pile on which they are standing.
- One can also encounter some piles whose height is greater than the pile they are standing on. So, they will need to remove some boxes from that pile to move forward. So, the task is to tell the total number of boxes needed to be removed from every pile(if necessary) during the journey to the ground.

The height of all the piles is given. Suppose that you are standing on the first pile. Print the total number of boxes to be removed.

**Examples**:

Input: arr[] = {3, 3, 2, 4, 1}

Output: 2

Explanation: After removing boxes, the heights of piles will be {3, 3, 2, 2, 1}

We are currently standing on 1st pile of height 3.

Step 1: We can move to the 2nd pile, since it’s height is equal to the height of the current pile.

Step 2: We can move to the 3rd pile of height 2, since it less than 3.

Step 3: We cannot go from 3rd pile to 4th pile(of height 4), so we need to remove 2 boxes from 4th pile to make it’s height equal to 2.

Step 4: We can easily move to the last pile since it’s height is 1 which is less than the height of the 4th pile of height 2(by removing 2 boxes in the previous step).

Input: arr[] = {5, 6, 7, 1}

Output: 3

Explanation: After removing boxes, the heights of piles will be {5, 5, 5, 1}

We are currently standing on 1st pile of height 5.

Step 1: We cannot move to the 2nd pile since it’s height is more. So, we remove 1 box and make its height equal to 5 and then we move forward.

Step 2: We cannot move to the 3rd pile of height 7, so we remove 2 boxes from it.

Step 3: We can easily move to the last pile since it’s height is 1 which is less than the height of the 3rd pile of height 5.

The idea is to traverse the array starting from left and every time before moving forward compare the height of the current pile with the previous pile. If the height of the current pile is greater than the previous pile, then increment count by the difference of the two heights otherwise move forward in the array.

Below is the implementation of the above approach:

## C++

`// C++ program to find the number of ` `// boxes to be removed ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find the number of ` `// boxes to be removed ` `int` `totalBoxesRemoved(` `int` `arr[], ` `int` `n) ` `{ ` ` ` `int` `count = 0; ` ` ` ` ` `// Store height of previous pile ` ` ` `int` `prev = arr[0]; ` ` ` ` ` `// Start traversing the array ` ` ` `for` `(` `int` `i = 1; i < n; i++) { ` ` ` `// if height of current pile is greater ` ` ` `// than previous pile ` ` ` `if` `(arr[i] > prev) { ` ` ` `// Increment count by difference ` ` ` `// of two heights ` ` ` `count += (arr[i] - prev); ` ` ` ` ` `// Update current height ` ` ` `arr[i] = prev; ` ` ` ` ` `// Update prev for next iteration ` ` ` `prev = arr[i]; ` ` ` `} ` ` ` `else` `{ ` ` ` `// Update prev for next iteration ` ` ` `prev = arr[i]; ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `count; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 5, 4, 7, 3, 2, 1 }; ` ` ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` ` ` `cout << totalBoxesRemoved(arr, n); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Java

`// Java program to find the number of ` `// boxes to be removed ` `import` `java.io.*; ` ` ` `class` `GFG { ` ` ` `// Function to find the number of ` `// boxes to be removed ` `static` `int` `totalBoxesRemoved(` `int` `arr[], ` `int` `n) ` `{ ` ` ` `int` `count = ` `0` `; ` ` ` ` ` `// Store height of previous pile ` ` ` `int` `prev = arr[` `0` `]; ` ` ` ` ` `// Start traversing the array ` ` ` `for` `(` `int` `i = ` `1` `; i < n; i++) { ` ` ` `// if height of current pile is greater ` ` ` `// than previous pile ` ` ` `if` `(arr[i] > prev) { ` ` ` `// Increment count by difference ` ` ` `// of two heights ` ` ` `count += (arr[i] - prev); ` ` ` ` ` `// Update current height ` ` ` `arr[i] = prev; ` ` ` ` ` `// Update prev for next iteration ` ` ` `prev = arr[i]; ` ` ` `} ` ` ` `else` `{ ` ` ` `// Update prev for next iteration ` ` ` `prev = arr[i]; ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `count; ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main (String[] args) { ` ` ` `int` `arr[] = { ` `5` `, ` `4` `, ` `7` `, ` `3` `, ` `2` `, ` `1` `}; ` ` ` ` ` `int` `n = arr.length; ` ` ` ` ` `System.out.println(totalBoxesRemoved(arr, n)); ` ` ` `} ` `} ` ` ` `// This code is contributed ` `// by inder_verma.. ` |

*chevron_right*

*filter_none*

## Python3

`# Python3 program to find the ` `# number of boxes to be removed ` ` ` `# Function to find the number ` `# of boxes to be removed ` `def` `totalBoxesRemoved(arr, n): ` ` ` ` ` `count ` `=` `0` ` ` ` ` `# Store height of previous pile ` ` ` `prev ` `=` `arr[` `0` `] ` ` ` ` ` `# Start traversing the array ` ` ` `for` `i ` `in` `range` `(` `1` `, n): ` ` ` ` ` `# if height of current pile ` ` ` `# is greater than previous pile ` ` ` `if` `(arr[i] > prev) : ` ` ` ` ` `# Increment count by ` ` ` `# difference of two heights ` ` ` `count ` `+` `=` `(arr[i] ` `-` `prev) ` ` ` ` ` `# Update current height ` ` ` `arr[i] ` `=` `prev ` ` ` ` ` `# Update prev for next ` ` ` `# iteration ` ` ` `prev ` `=` `arr[i] ` ` ` ` ` `else` `: ` ` ` `# Update prev for next ` ` ` `# iteration ` ` ` `prev ` `=` `arr[i] ` ` ` ` ` `return` `count ` ` ` `# Driver code ` `arr ` `=` `[ ` `5` `, ` `4` `, ` `7` `, ` `3` `, ` `2` `, ` `1` `] ` ` ` `n ` `=` `len` `(arr) ` ` ` `print` `(totalBoxesRemoved(arr, n)) ` ` ` `# This code is contributed ` `# by Yatin Gupta ` |

*chevron_right*

*filter_none*

## C#

`// C# program to find the number of ` `// boxes to be removed ` `using` `System; ` ` ` `class` `GFG { ` ` ` `// Function to find the number of ` `// boxes to be removed ` `static` `int` `totalBoxesRemoved(` `int` `[]arr, ` `int` `n) ` `{ ` ` ` `int` `count = 0; ` ` ` ` ` `// Store height of previous pile ` ` ` `int` `prev = arr[0]; ` ` ` ` ` `// Start traversing the array ` ` ` `for` `(` `int` `i = 1; i < n; i++) { ` ` ` `// if height of current pile is greater ` ` ` `// than previous pile ` ` ` `if` `(arr[i] > prev) { ` ` ` `// Increment count by difference ` ` ` `// of two heights ` ` ` `count += (arr[i] - prev); ` ` ` ` ` `// Update current height ` ` ` `arr[i] = prev; ` ` ` ` ` `// Update prev for next iteration ` ` ` `prev = arr[i]; ` ` ` `} ` ` ` `else` `{ ` ` ` `// Update prev for next iteration ` ` ` `prev = arr[i]; ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `count; ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `Main () { ` ` ` `int` `[]arr = { 5, 4, 7, 3, 2, 1 }; ` ` ` ` ` `int` `n = arr.Length; ` ` ` ` ` `Console.WriteLine(totalBoxesRemoved(arr, n)); ` ` ` `} ` `} ` ` ` `// This code is contributed ` `// by shs ` |

*chevron_right*

*filter_none*

## PHP

`<?php ` `// PHP program to find the number ` `// of boxes to be removed ` ` ` ` ` `// Function to find the number ` `// of boxes to be removed ` `function` `totalBoxesRemoved(` `$arr` `, ` `$n` `) ` `{ ` ` ` `$count` `= 0; ` ` ` ` ` `// Store height of previous pile ` ` ` `$prev` `= ` `$arr` `[0]; ` ` ` ` ` `// Start traversing the array ` ` ` `for` `(` `$i` `= 1; ` `$i` `<` `$n` `; ` `$i` `++) ` ` ` `{ ` ` ` `// if height of current pile is ` ` ` `// greater than previous pile ` ` ` `if` `(` `$arr` `[` `$i` `] > ` `$prev` `) ` ` ` `{ ` ` ` `// Increment count by difference ` ` ` `// of two heights ` ` ` `$count` `+= (` `$arr` `[` `$i` `] - ` `$prev` `); ` ` ` ` ` `// Update current height ` ` ` `$arr` `[` `$i` `] = ` `$prev` `; ` ` ` ` ` `// Update prev for next iteration ` ` ` `$prev` `= ` `$arr` `[` `$i` `]; ` ` ` `} ` ` ` `else` ` ` `{ ` ` ` `// Update prev for next iteration ` ` ` `$prev` `= ` `$arr` `[` `$i` `]; ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `$count` `; ` `} ` ` ` `// Driver code ` `$arr` `= ` `array` `( 5, 4, 7, 3, 2, 1 ); ` ` ` `$n` `= ` `count` `(` `$arr` `); ` ` ` `echo` `totalBoxesRemoved(` `$arr` `, ` `$n` `); ` ` ` `// This code is contributed ` `// by shs ` `?> ` |

*chevron_right*

*filter_none*

**Output:**

3

**Time Complexity**: O(N), where N is the total number of piles.

## Recommended Posts:

- Find the minimum number of elements that should be removed to make an array good
- Number of visible boxes after putting one inside another
- Minimum number of stacks possible using boxes of given capacities
- Find the position of the last removed element from the array
- Color N boxes using M colors such that K boxes have different color from the box on its left
- Minimum number of elements to be removed such that the sum of the remaining elements is equal to k
- Minimum boxes required to carry all gifts
- Minimum time required to transport all the boxes from source to the destination under the given constraints
- Length of Longest sub-string that can be removed
- Minimum Circles needed to be removed so that all remaining circles are non intersecting
- Find a number which give minimum sum when XOR with every number of array of integers
- Minimum elements to be removed such that sum of adjacent elements is always odd
- Minimum elements to be removed such that sum of adjacent elements is always even
- Find the Number Occurring Odd Number of Times
- Find number of subarrays with even sum

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.