Given an array **arr[]** of size **N**, the task is to print the distance of every array element from its next greater element. For array elements having no next greater element, print 0.

**Examples:**

Input:arr[] = {73, 74, 75, 71, 69, 72, 76, 73}Output:{1, 1, 4, 2, 1, 1, 0, 0}Explanation:

The next greater element for 73 is 74, which is at position 1. Distance = 1 – 0 = 1

The next greater element for 74 is 75, which is at position 2. Distance = 2 – 1 = 1

The next greater element for 75 is 76, which is at position 6. Distance = 6 – 2 = 4

The next greater element for 71 is 72, which is at position 5. Distance = 5 – 3 = 2

The next greater element for 69 is 72, which is at position 5. Distance = 5 – 4 = 1

The next greater element for 72 is 76, which is at position 6. Distance = 6 – 5 = 1

No, next greater element for 76. Distance = 0

No, next greater element for 73. Distance = 0

Input:arr[] = {5, 4, 3, 2, 1}Output:{0, 0, 0, 0, 0}

**Naive Approach: **The simplest approach is to traverse the array and for every array element, traverse to its right to obtain its next greater element and calculate the difference between the indices.

**Time Complexity:** O(N^{2}) **Auxiliary Space:** O(1)

**Efficient Approach: **To optimize the above approach, the idea is to use Stack to find the next greater element.

Below are the steps:

- Maintain a
**Stack**which will contain the elements in non-increasing order. - Check if the current element
**arr[i]**is greater than the element at the top of the stack. - Keep popping all the elements from the stack one by one from the top, that are found to be smaller than
**arr[i]**and calculate the distance for each of them as the difference of current index and the index of the popped element. - Push the current element into the stack and repeat the above steps.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the ` `// above approach ` `#include<bits/stdc++.h>` `using` `namespace` `std;` `vector<` `int` `> mindistance(vector<` `int` `> arr)` `{` ` ` `int` `N = arr.size();` ` ` ` ` `// Stores the required distances ` ` ` `vector<` `int` `> ans(N);` ` ` `int` `st = 0;` ` ` ` ` `// Maintain a stack of elements ` ` ` `// in non-increasing order ` ` ` `for` `(` `int` `i = 0; i < N - 1; i++)` ` ` `{` ` ` `if` `(arr[i] < arr[i + 1])` ` ` `{` ` ` `ans[i] = 1;` ` ` `}` ` ` `else` ` ` `{` ` ` `st = i + 1;` ` ` `while` `(st <= N - 1)` ` ` `{` ` ` `if` `(arr[i] < arr[st]) ` ` ` `{` ` ` `ans[i] = st - i;` ` ` `break` `;` ` ` `}` ` ` `else` ` ` `{` ` ` `st++;` ` ` `}` ` ` `}` ` ` `}` ` ` `}` ` ` `return` `ans;` `}` `// Driver code` `int` `main()` `{` ` ` `vector<` `int` `> arr = { 73, 74, 75, 71, ` ` ` `69, 72, 76, 73 };` ` ` ` ` `vector<` `int` `> x = mindistance(arr);` ` ` ` ` `cout << ` `"["` `;` ` ` `for` `(` `int` `i = 0; i < x.size(); i++)` ` ` `{` ` ` `if` `(i == x.size() - 1)` ` ` `cout << x[i];` ` ` `else` ` ` `cout << x[i] << ` `", "` `;` ` ` `}` ` ` `cout << ` `"]"` `;` `}` `// This code is contributed by SURENDRA_GANGWAR` |

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## Java

`// Java implementation of the ` `// above approach ` `import` `java.io.*;` `class` `GFG{` ` ` `public` `static` `int` `[] mindistance(` `int` `[] arr)` `{` ` ` `int` `N = arr.length;` ` ` ` ` `// Stores the required distances ` ` ` `int` `[] ans = ` `new` `int` `[N];` ` ` `int` `st = ` `0` `;` ` ` ` ` `// Maintain a stack of elements ` ` ` `// in non-increasing order ` ` ` `for` `(` `int` `i = ` `0` `; i < N - ` `1` `; i++)` ` ` `{` ` ` `if` `(arr[i] < arr[i + ` `1` `])` ` ` `{` ` ` `ans[i] = ` `1` `;` ` ` `}` ` ` `else` ` ` `{` ` ` `st = i + ` `1` `;` ` ` `while` `(st <= N - ` `1` `)` ` ` `{` ` ` `if` `(arr[i] < arr[st]) ` ` ` `{` ` ` `ans[i] = st - i;` ` ` `break` `;` ` ` `}` ` ` `else` ` ` `{` ` ` `st++;` ` ` `}` ` ` `}` ` ` `}` ` ` `}` ` ` `return` `ans;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `arr[] = ` `new` `int` `[]{ ` `73` `, ` `74` `, ` `75` `, ` `71` `, ` ` ` `69` `, ` `72` `, ` `76` `, ` `73` `};` ` ` ` ` `int` `x[] = mindistance(arr);` ` ` ` ` `System.out.print(` `"["` `);` ` ` `for` `(` `int` `i = ` `0` `; i < x.length; i++)` ` ` `System.out.print(x[i]+` `", "` `);` ` ` ` ` `System.out.print(` `"]"` `);` `}` `}` `// This code is contributed by sai-sampath mahajan` `// and ramprasad kondoju` |

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## Python3

`# Python3 implementation of the` `# above approach` `def` `mindistance(arr, N): ` ` ` ` ` `if` `N <` `=` `1` `:` ` ` `return` `[` `0` `]` ` ` ` ` `# Stores the required distances` ` ` `ans ` `=` `[` `0` `for` `i ` `in` `range` `(N)]` ` ` `st ` `=` `[` `0` `]` ` ` ` ` `# Maintain a stack of elements` ` ` `# in non-increasing order` ` ` `for` `i ` `in` `range` `(` `1` `, N): ` ` ` ` ` `# If the current element exceeds ` ` ` `# the element at the top of the stack` ` ` `while` `(st ` `and` `arr[i] > arr[st[` `-` `1` `]]): ` ` ` `pos ` `=` `st.pop()` ` ` `ans[pos] ` `=` `i ` `-` `pos` ` ` ` ` `# Push the current index to the stack` ` ` `st.append(i)` ` ` `return` `ans` `# Given array` `arr ` `=` `[` `73` `, ` `74` `, ` `75` `, ` `71` `, ` `69` `, ` `72` `, ` `76` `, ` `73` `]` `N ` `=` `len` `(arr)` `# Function call` `print` `(mindistance(arr, N))` |

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**Output:**

[1, 1, 4, 2, 1, 1, 0, 0]

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

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