# Replace elements with absolute difference of smallest element on left and largest element on right

Given an array arr[] of N integers. The task is to replace all the elements of the array by the absolute difference of the smallest element on its left and the largest element on its right.

Examples:

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

Element Smallest on its left Largest on its right Absolute difference
1 NULL 5 5
5 1 4 3
2 1 4 3
4 1 3 2
3 1 NULL 1

Input: arr[] = {4, 3, 6, 2, 1, 20, 9, 10, 15, 6}
Output: 20 16 17 17 18 14 14 14 5 1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive approach: For every element arr[i] of the array, find the minimum element in the subarray arr[0…i-1] then the maximum element in the subarray arr[i+1…n-1] and print the absolute difference between the two. The time complexity of this approach will be O(N2).

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Utility function to print the ` `// elements of an array ` `void` `printArray(``int` `arr[], ``int` `n) ` `{ ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``cout << arr[i] << ``" "``; ` `    ``} ` `} ` ` `  `// Function to return the minimum element ` `// in the subarray arr[i...j] ` `int` `getMin(``int` `arr[], ``int` `i, ``int` `j) ` `{ ` ` `  `    ``// To store the minimum element ` `    ``int` `minVal = arr[i++]; ` `    ``while` `(i <= j) { ` ` `  `        ``// Update the minimum element so far ` `        ``minVal = min(minVal, arr[i]); ` `        ``i++; ` `    ``} ` ` `  `    ``// Return the minimum element found ` `    ``return` `minVal; ` `} ` ` `  `// Function to return the maximum element ` `// in the subarray arr[i...j] ` `int` `getMax(``int` `arr[], ``int` `i, ``int` `j) ` `{ ` ` `  `    ``// To store the maximum element ` `    ``int` `maxVal = arr[i++]; ` `    ``while` `(i <= j) { ` ` `  `        ``// Update the maximum element so far ` `        ``maxVal = max(maxVal, arr[i]); ` `        ``i++; ` `    ``} ` ` `  `    ``// Return the maximum element found ` `    ``return` `maxVal; ` `} ` ` `  `// Function to generate the array ` `// with the given operations ` `void` `generateArr(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// Base cases ` `    ``if` `(n == 0) ` `        ``return``; ` `    ``if` `(n == 1) { ` `        ``cout << arr[0]; ` `        ``return``; ` `    ``} ` ` `  `    ``// To store the new array elements ` `    ``int` `tmpArr[n]; ` ` `  `    ``// The first element has no ` `    ``// element on its left ` `    ``tmpArr[0] = getMax(arr, 1, n - 1); ` ` `  `    ``// From the second element to the ` `    ``// second last element ` `    ``for` `(``int` `i = 1; i < n - 1; i++) { ` ` `  `        ``// Absolute difference of the maximum ` `        ``// element to the right and the ` `        ``// minimum element to the left ` `        ``tmpArr[i] = ``abs``(getMax(arr, i + 1, n - 1) ` `                        ``- getMin(arr, 0, i - 1)); ` `    ``} ` ` `  `    ``// The last element has no ` `    ``// element on its right ` `    ``tmpArr[n - 1] = getMin(arr, 0, n - 2); ` ` `  `    ``// Print the generated array ` `    ``printArray(tmpArr, n); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 5, 2, 4, 3 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``generateArr(arr, n); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of the approach  ` `class` `GFG ` `{ ` `     `  `// Utility function to print the  ` `// elements of an array  ` `static` `void` `printArray(``int` `arr[], ``int` `n)  ` `{  ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{  ` `        ``System.out.print(arr[i] + ``" "``);  ` `    ``}  ` `}  ` ` `  `// Function to return the minimum element  ` `// in the subarray arr[i...j]  ` `static` `int` `getMin(``int` `arr[], ``int` `i, ``int` `j)  ` `{  ` ` `  `    ``// To store the minimum element  ` `    ``int` `minVal = arr[i++];  ` `    ``while` `(i <= j)  ` `    ``{  ` ` `  `        ``// Update the minimum element so far  ` `        ``minVal = Math.min(minVal, arr[i]);  ` `        ``i++;  ` `    ``}  ` ` `  `    ``// Return the minimum element found  ` `    ``return` `minVal;  ` `}  ` ` `  `// Function to return the maximum element  ` `// in the subarray arr[i...j]  ` `static` `int` `getMax(``int` `arr[], ``int` `i, ``int` `j)  ` `{  ` ` `  `    ``// To store the maximum element  ` `    ``int` `maxVal = arr[i++];  ` `    ``while` `(i <= j)  ` `    ``{  ` ` `  `        ``// Update the maximum element so far  ` `        ``maxVal = Math.max(maxVal, arr[i]);  ` `        ``i++;  ` `    ``}  ` ` `  `    ``// Return the maximum element found  ` `    ``return` `maxVal;  ` `}  ` ` `  `// Function to generate the array  ` `// with the given operations  ` `static` `void` `generateArr(``int` `arr[], ``int` `n)  ` `{  ` ` `  `    ``// Base cases  ` `    ``if` `(n == ``0``)  ` `        ``return``;  ` `    ``if` `(n == ``1``) ` `    ``{  ` `        ``System.out.println(arr[``0``]);  ` `        ``return``;  ` `    ``}  ` ` `  `    ``// To store the new array elements  ` `    ``int` `tmpArr[] = ``new` `int``[n];  ` ` `  `    ``// The first element has no  ` `    ``// element on its left  ` `    ``tmpArr[``0``] = getMax(arr, ``1``, n - ``1``);  ` ` `  `    ``// From the second element to the  ` `    ``// second last element  ` `    ``for` `(``int` `i = ``1``; i < n - ``1``; i++)  ` `    ``{  ` ` `  `        ``// Absolute difference of the maximum  ` `        ``// element to the right and the  ` `        ``// minimum element to the left  ` `        ``tmpArr[i] = Math.abs(getMax(arr, i + ``1``, n - ``1``) -  ` `                             ``getMin(arr, ``0``, i - ``1``));  ` `    ``}  ` ` `  `    ``// The last element has no  ` `    ``// element on its right  ` `    ``tmpArr[n - ``1``] = getMin(arr, ``0``, n - ``2``);  ` ` `  `    ``// Print the generated array  ` `    ``printArray(tmpArr, n);  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main (String[] args)  ` `{  ` `    ``int` `arr[] = { ``1``, ``5``, ``2``, ``4``, ``3` `};  ` `    ``int` `n = arr.length;  ` ` `  `    ``generateArr(arr, n);  ` `} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

 `# Python3 implementation of the approach ` ` `  `# Utility function to prthe ` `# elements of an array ` `def` `printArray(arr, n): ` `    ``for` `i ``in` `range``(n): ` `        ``print``(arr[i], end ``=` `" "``) ` ` `  `# Function to return the minimum element ` `# in the subarray arr[i...j] ` `def` `getMin(arr, i, j): ` ` `  `    ``# To store the minimum element ` `    ``minVal ``=` `arr[i] ` `    ``i ``+``=` `1` `    ``while` `(i <``=` `j): ` ` `  `        ``# Update the minimum element so far ` `        ``minVal ``=` `min``(minVal, arr[i]) ` `        ``i ``+``=` `1` ` `  `    ``# Return the minimum element found ` `    ``return` `minVal ` ` `  `# Function to return the maximum element ` `# in the subarray arr[i...j] ` `def` `getMax(arr, i, j): ` ` `  `    ``# To store the maximum element ` `    ``maxVal ``=` `arr[i] ` `    ``i ``+``=` `1` `    ``while` `(i <``=` `j): ` ` `  `        ``# Update the maximum element so far ` `        ``maxVal ``=` `max``(maxVal, arr[i]) ` `        ``i ``+``=` `1` ` `  `    ``# Return the maximum element found ` `    ``return` `maxVal ` ` `  `# Function to generate the array ` `# With the given operations ` `def` `generateArr(arr, n): ` ` `  `    ``# Base cases ` `    ``if` `(n ``=``=` `0``): ` `        ``return` `    ``if` `(n ``=``=` `1``): ` `        ``print``(arr[``0``], end ``=` `"") ` `        ``return` ` `  `    ``# To store the new array elements ` `    ``tmpArr ``=` `[``0` `for` `i ``in` `range``(n)] ` ` `  `    ``# The first element has no ` `    ``# element on its left ` `    ``tmpArr[``0``] ``=` `getMax(arr, ``1``, n ``-` `1``) ` ` `  `    ``# From the second element to the ` `    ``# second last element ` `    ``for` `i ``in` `range``(``1``, n ``-` `1``): ` ` `  `        ``# Absolute difference of the maximum ` `        ``# element to the right and the ` `        ``# minimum element to the left ` `        ``tmpArr[i] ``=` `abs``(getMax(arr, i ``+` `1``, n ``-` `1``) ``-` `                        ``getMin(arr, ``0``, i ``-` `1``)) ` ` `  `    ``# The last element has no ` `    ``# element on its right ` `    ``tmpArr[n ``-` `1``] ``=` `getMin(arr, ``0``, n ``-` `2``) ` ` `  `    ``# Print the generated array ` `    ``printArray(tmpArr, n) ` ` `  `# Driver code ` `arr ``=` `[``1``, ``5``, ``2``, ``4``, ``3``] ` `n ``=` `len``(arr) ` ` `  `generateArr(arr, n) ` ` `  `# This code is contributed by Mohit Kumar `

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG ` `{ ` `         `  `// Utility function to print the  ` `// elements of an array  ` `static` `void` `printArray(``int` `[]arr, ``int` `n)  ` `{  ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{  ` `        ``Console.Write(arr[i] + ``" "``);  ` `    ``}  ` `}  ` ` `  `// Function to return the minimum element  ` `// in the subarray arr[i...j]  ` `static` `int` `getMin(``int` `[]arr, ``int` `i, ``int` `j)  ` `{  ` ` `  `    ``// To store the minimum element  ` `    ``int` `minVal = arr[i++];  ` `    ``while` `(i <= j)  ` `    ``{  ` ` `  `        ``// Update the minimum element so far  ` `        ``minVal = Math.Min(minVal, arr[i]);  ` `        ``i++;  ` `    ``}  ` ` `  `    ``// Return the minimum element found  ` `    ``return` `minVal;  ` `}  ` ` `  `// Function to return the maximum element  ` `// in the subarray arr[i...j]  ` `static` `int` `getMax(``int` `[]arr, ``int` `i, ``int` `j)  ` `{  ` ` `  `    ``// To store the maximum element  ` `    ``int` `maxVal = arr[i++];  ` `    ``while` `(i <= j)  ` `    ``{  ` ` `  `        ``// Update the maximum element so far  ` `        ``maxVal = Math.Max(maxVal, arr[i]);  ` `        ``i++;  ` `    ``}  ` ` `  `    ``// Return the maximum element found  ` `    ``return` `maxVal;  ` `}  ` ` `  `// Function to generate the array  ` `// with the given operations  ` `static` `void` `generateArr(``int` `[]arr, ``int` `n)  ` `{  ` ` `  `    ``// Base cases  ` `    ``if` `(n == 0)  ` `        ``return``;  ` `    ``if` `(n == 1) ` `    ``{  ` `        ``Console.WriteLine(arr[0]);  ` `        ``return``;  ` `    ``}  ` ` `  `    ``// To store the new array elements  ` `    ``int` `[]tmpArr = ``new` `int``[n];  ` ` `  `    ``// The first element has no  ` `    ``// element on its left  ` `    ``tmpArr[0] = getMax(arr, 1, n - 1);  ` ` `  `    ``// From the second element to the  ` `    ``// second last element  ` `    ``for` `(``int` `i = 1; i < n - 1; i++)  ` `    ``{  ` ` `  `        ``// Absolute difference of the maximum  ` `        ``// element to the right and the  ` `        ``// minimum element to the left  ` `        ``tmpArr[i] = Math.Abs(getMax(arr, i + 1, n - 1) -  ` `                             ``getMin(arr, 0, i - 1));  ` `    ``}  ` ` `  `    ``// The last element has no  ` `    ``// element on its right  ` `    ``tmpArr[n - 1] = getMin(arr, 0, n - 2);  ` ` `  `    ``// Print the generated array  ` `    ``printArray(tmpArr, n);  ` `}  ` ` `  `// Driver code  ` `static` `public` `void` `Main () ` `{ ` `    ``int` `[]arr = { 1, 5, 2, 4, 3 };  ` `    ``int` `n = arr.Length;  ` ` `  `    ``generateArr(arr, n);  ` `} ` `} ` ` `  `// This code is contributed by ajit. `

Output:
```5 3 3 2 1
```

Efficient approach: Create an array suffixMax[] where suffixMax[i] will store the maximum element in the subarray arr[i…N-1]. Also take a variable minSoFar which will store the minimum element so far on traversing the array from left to right. Now, for every element arr[i] the updated value will be abs(suffixMax[i + 1] – minSoFar).

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Utility function to print the ` `// elements of an array ` `void` `printArray(``int` `arr[], ``int` `n) ` `{ ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``cout << arr[i] << ``" "``; ` `    ``} ` `} ` ` `  `// Function to return the minimum element ` `// in the subarray arr[i...j] ` `int` `getMin(``int` `arr[], ``int` `i, ``int` `j) ` `{ ` ` `  `    ``// To store the minimum element ` `    ``int` `minVal = arr[i++]; ` `    ``while` `(i <= j) { ` ` `  `        ``// Update the minimum element so far ` `        ``minVal = min(minVal, arr[i]); ` `        ``i++; ` `    ``} ` ` `  `    ``// Return the minimum element found ` `    ``return` `minVal; ` `} ` ` `  `// Function to return the maximum element ` `// in the subarray arr[i...j] ` `int` `getMax(``int` `arr[], ``int` `i, ``int` `j) ` `{ ` ` `  `    ``// To store the maximum element ` `    ``int` `maxVal = arr[i++]; ` `    ``while` `(i <= j) { ` ` `  `        ``// Update the maximum element so far ` `        ``maxVal = max(maxVal, arr[i]); ` `        ``i++; ` `    ``} ` ` `  `    ``// Return the maximum element found ` `    ``return` `maxVal; ` `} ` ` `  `// Function to generate the array ` `// with the given operations ` `void` `generateArr(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// Base cases ` `    ``if` `(n == 0) ` `        ``return``; ` `    ``if` `(n == 1) { ` `        ``cout << arr[0]; ` `        ``return``; ` `    ``} ` ` `  `    ``// To suffixMax[i] will store the maximum ` `    ``// element in the subarray arr[i...n-1] ` `    ``int` `suffixMax[n]; ` `    ``suffixMax[n - 1] = arr[n - 1]; ` `    ``for` `(``int` `i = n - 2; i >= 0; i--) ` `        ``suffixMax[i] = max(arr[i], suffixMax[i + 1]); ` ` `  `    ``// To store the minimum element on the left ` `    ``int` `minSoFar = arr[0]; ` ` `  `    ``// The first element has no ` `    ``// element on its left ` `    ``arr[0] = suffixMax[1]; ` ` `  `    ``// From the second element to the ` `    ``// second last element ` `    ``for` `(``int` `i = 1; i < n - 1; i++) { ` ` `  `        ``// Store a copy of the currene element ` `        ``int` `temp = arr[i]; ` ` `  `        ``// Absolute difference of the maximum ` `        ``// element to the right and the ` `        ``// minimum element to the left ` `        ``arr[i] = ``abs``(suffixMax[i + 1] - minSoFar); ` ` `  `        ``// Update the minimum element so far ` `        ``minSoFar = min(minSoFar, temp); ` `    ``} ` ` `  `    ``// The last element has no ` `    ``// element on its right ` `    ``arr[n - 1] = minSoFar; ` ` `  `    ``// Print the updated array ` `    ``printArray(arr, n); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 5, 2, 4, 3 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``generateArr(arr, n); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of the approach ` `class` `GFG ` `{ ` ` `  `// Utility function to print the ` `// elements of an array ` `static` `void` `printArray(``int` `arr[], ``int` `n) ` `{ ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` `        ``System.out.print(arr[i] + ``" "``); ` `    ``} ` `} ` ` `  `// Function to return the minimum element ` `// in the subarray arr[i...j] ` `static` `int` `getMin(``int` `arr[], ``int` `i, ``int` `j) ` `{ ` ` `  `    ``// To store the minimum element ` `    ``int` `minVal = arr[i++]; ` `    ``while` `(i <= j)  ` `    ``{ ` ` `  `        ``// Update the minimum element so far ` `        ``minVal = Math.min(minVal, arr[i]); ` `        ``i++; ` `    ``} ` ` `  `    ``// Return the minimum element found ` `    ``return` `minVal; ` `} ` ` `  `// Function to return the maximum element ` `// in the subarray arr[i...j] ` `static` `int` `getMax(``int` `arr[], ``int` `i, ``int` `j) ` `{ ` ` `  `    ``// To store the maximum element ` `    ``int` `maxVal = arr[i++]; ` `    ``while` `(i <= j) ` `    ``{ ` ` `  `        ``// Update the maximum element so far ` `        ``maxVal = Math.max(maxVal, arr[i]); ` `        ``i++; ` `    ``} ` ` `  `    ``// Return the maximum element found ` `    ``return` `maxVal; ` `} ` ` `  `// Function to generate the array ` `// with the given operations ` `static` `void` `generateArr(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// Base cases ` `    ``if` `(n == ``0``) ` `        ``return``; ` `    ``if` `(n == ``1``)  ` `    ``{ ` `        ``System.out.print(arr[``0``]); ` `        ``return``; ` `    ``} ` ` `  `    ``// To suffixMax[i] will store the maximum ` `    ``// element in the subarray arr[i...n-1] ` `    ``int` `[]suffixMax = ``new` `int``[n]; ` `    ``suffixMax[n - ``1``] = arr[n - ``1``]; ` `    ``for` `(``int` `i = n - ``2``; i >= ``0``; i--) ` `        ``suffixMax[i] = Math.max(arr[i],  ` `                                ``suffixMax[i + ``1``]); ` ` `  `    ``// To store the minimum element on the left ` `    ``int` `minSoFar = arr[``0``]; ` ` `  `    ``// The first element has no ` `    ``// element on its left ` `    ``arr[``0``] = suffixMax[``1``]; ` ` `  `    ``// From the second element to the ` `    ``// second last element ` `    ``for` `(``int` `i = ``1``; i < n - ``1``; i++)  ` `    ``{ ` ` `  `        ``// Store a copy of the currene element ` `        ``int` `temp = arr[i]; ` ` `  `        ``// Absolute difference of the maximum ` `        ``// element to the right and the ` `        ``// minimum element to the left ` `        ``arr[i] = Math.abs(suffixMax[i + ``1``] -  ` `                                 ``minSoFar); ` ` `  `        ``// Update the minimum element so far ` `        ``minSoFar = Math.min(minSoFar, temp); ` `    ``} ` ` `  `    ``// The last element has no ` `    ``// element on its right ` `    ``arr[n - ``1``] = minSoFar; ` ` `  `    ``// Print the updated array ` `    ``printArray(arr, n); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main (String[] args) ` `{ ` `    ``int` `arr[] = { ``1``, ``5``, ``2``, ``4``, ``3` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``generateArr(arr, n); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

 `// C# implementation for above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Utility function to print the ` `// elements of an array ` `static` `void` `printArray(``int` `[]arr, ``int` `n) ` `{ ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``Console.Write(arr[i] + ``" "``); ` `    ``} ` `} ` ` `  `// Function to return the minimum element ` `// in the subarray arr[i...j] ` `static` `int` `getMin(``int` `[]arr, ``int` `i, ``int` `j) ` `{ ` ` `  `    ``// To store the minimum element ` `    ``int` `minVal = arr[i++]; ` `    ``while` `(i <= j)  ` `    ``{ ` ` `  `        ``// Update the minimum element so far ` `        ``minVal = Math.Min(minVal, arr[i]); ` `        ``i++; ` `    ``} ` ` `  `    ``// Return the minimum element found ` `    ``return` `minVal; ` `} ` ` `  `// Function to return the maximum element ` `// in the subarray arr[i...j] ` `static` `int` `getMax(``int` `[]arr, ``int` `i, ``int` `j) ` `{ ` ` `  `    ``// To store the maximum element ` `    ``int` `maxVal = arr[i++]; ` `    ``while` `(i <= j) ` `    ``{ ` ` `  `        ``// Update the maximum element so far ` `        ``maxVal = Math.Max(maxVal, arr[i]); ` `        ``i++; ` `    ``} ` ` `  `    ``// Return the maximum element found ` `    ``return` `maxVal; ` `} ` ` `  `// Function to generate the array ` `// with the given operations ` `static` `void` `generateArr(``int` `[]arr, ``int` `n) ` `{ ` ` `  `    ``// Base cases ` `    ``if` `(n == 0) ` `        ``return``; ` `    ``if` `(n == 1)  ` `    ``{ ` `        ``Console.Write(arr[0]); ` `        ``return``; ` `    ``} ` ` `  `    ``// To suffixMax[i] will store the maximum ` `    ``// element in the subarray arr[i...n-1] ` `    ``int` `[]suffixMax = ``new` `int``[n]; ` `    ``suffixMax[n - 1] = arr[n - 1]; ` `    ``for` `(``int` `i = n - 2; i >= 0; i--) ` `        ``suffixMax[i] = Math.Max(arr[i],  ` `                                ``suffixMax[i + 1]); ` ` `  `    ``// To store the minimum element on the left ` `    ``int` `minSoFar = arr[0]; ` ` `  `    ``// The first element has no ` `    ``// element on its left ` `    ``arr[0] = suffixMax[1]; ` ` `  `    ``// From the second element to the ` `    ``// second last element ` `    ``for` `(``int` `i = 1; i < n - 1; i++)  ` `    ``{ ` ` `  `        ``// Store a copy of the currene element ` `        ``int` `temp = arr[i]; ` ` `  `        ``// Absolute difference of the maximum ` `        ``// element to the right and the ` `        ``// minimum element to the left ` `        ``arr[i] = Math.Abs(suffixMax[i + 1] -  ` `                                 ``minSoFar); ` ` `  `        ``// Update the minimum element so far ` `        ``minSoFar = Math.Min(minSoFar, temp); ` `    ``} ` ` `  `    ``// The last element has no ` `    ``// element on its right ` `    ``arr[n - 1] = minSoFar; ` ` `  `    ``// Print the updated array ` `    ``printArray(arr, n); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main (String[] args) ` `{ ` `    ``int` `[]arr = { 1, 5, 2, 4, 3 }; ` `    ``int` `n = arr.Length; ` ` `  `    ``generateArr(arr, n); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:
```5 3 3 2 1
```

Check out this Author's contributed articles.

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.

Article Tags :
Practice Tags :