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
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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Utility function to print the// elements of an arrayvoid 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 operationsvoid 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 codeint main(){ int arr[] = { 1, 5, 2, 4, 3 }; int n = sizeof(arr) / sizeof(arr[0]); generateArr(arr, n); return 0;} |
Java
// 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
# Python3 implementation of the approach# Utility function to print the# elements of an arraydef 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 operationsdef 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 codearr = [1, 5, 2, 4, 3]n = len(arr)generateArr(arr, n)# This code is contributed by Mohit Kumar |
C#
// 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. |
Javascript
<script> // JavaScript implementation of the approach // Utility function to print the // elements of an array function printArray(arr, n) { for (let i = 0; i < n; i++) { document.write(arr[i] + " "); } } // Function to return the minimum element // in the subarray arr[i...j] function getMin(arr, i, j) { // To store the minimum element let 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] function getMax(arr, i, j) { // To store the maximum element let 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 function generateArr(arr, n) { // Base cases if (n == 0) return; if (n == 1) { document.write(arr[0] + "</br>"); return; } // To store the new array elements let tmpArr = new Array(n); tmpArr.fill(0); // 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 (let 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); } let arr = [ 1, 5, 2, 4, 3 ]; let n = arr.length; generateArr(arr, n); // This code is contributed by suresh07.</script> |
Output:
5 3 3 2 1
Time complexity: O(N2) where N is the size of the given array
Auxiliary space: O(N)
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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Utility function to print the// elements of an arrayvoid printArray(int arr[], int n){ for (int i = 0; i < n; i++) { cout << arr[i] << " "; }}// Function to generate the array// with the given operationsvoid 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 codeint main(){ int arr[] = { 1, 5, 2, 4, 3 }; int n = sizeof(arr) / sizeof(arr[0]); generateArr(arr, n); return 0;} |
Java
// Java implementation of the approachclass GFG{// Utility function to print the// elements of an arraystatic void printArray(int arr[], int n){ for (int i = 0; i < n; i++) { System.out.print(arr[i] + " "); }}// Function to generate the array// with the given operationsstatic 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 codepublic 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 |
Python3
# Python 3 implementation of the approach # Utility function to print the# elements of an arraydef printArray(arr, n): for i in range( n): print(arr[i],end= " ") # Function to generate the array# with the given operationsdef generateArr(arr, n): # Base cases if (n == 0): return if (n == 1): print( arr[0]) return # To suffixMax[i] will store the maximum # element in the subarray arr[i...n-1] suffixMax=[0]*n suffixMax[n - 1] = arr[n - 1] for i in range(n - 2, -1 ,-1): suffixMax[i] = max(arr[i], suffixMax[i + 1]) # To store the minimum element on the left 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 i in range( 1,n - 1): # Store a copy of the currene element 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 codeif __name__ == "__main__": arr = [ 1, 5, 2, 4, 3 ] n = len(arr) generateArr(arr, n)# This code is contributed by chitranayal |
C#
// C# implementation for above approachusing System;class GFG{// Utility function to print the// elements of an arraystatic void printArray(int []arr, int n){ for (int i = 0; i < n; i++) { Console.Write(arr[i] + " "); }}// Function to generate the array// with the given operationsstatic 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 codepublic 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 |
Javascript
<script> // JavaScript implementation for above approach // Utility function to print the // elements of an array function printArray(arr, n) { for (let i = 0; i < n; i++) { document.write(arr[i] + " "); } } // Function to generate the array // with the given operations function generateArr(arr, n) { // Base cases if (n == 0) return; if (n == 1) { document.write(arr[0]); return; } // To suffixMax[i] will store the maximum // element in the subarray arr[i...n-1] let suffixMax = new Array(n); suffixMax.fill(0); suffixMax[n - 1] = arr[n - 1]; for (let i = n - 2; i >= 0; i--) suffixMax[i] = Math.max(arr[i], suffixMax[i + 1]); // To store the minimum element on the left let 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 (let i = 1; i < n - 1; i++) { // Store a copy of the currene element let 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); } let arr = [ 1, 5, 2, 4, 3 ]; let n = arr.length; generateArr(arr, n); </script> |
Output:
5 3 3 2 1
Time complexity: O(N) where N is size of given array
Auxiliary space: O(N)
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