Replace every array element by multiplication of previous and next - GeeksQuiz
Given an array of integers, update every element with multiplication of previous and next elements with following exceptions.
a) First element is replaced by multiplication of first and second.
b) Last element is replaced by multiplication of last and second last.
Read full article from Replace every array element by multiplication of previous and next - GeeksQuiz
Given an array of integers, update every element with multiplication of previous and next elements with following exceptions.
a) First element is replaced by multiplication of first and second.
b) Last element is replaced by multiplication of last and second last.
Input: arr[] = {2, 3, 4, 5, 6}
Output: arr[] = {6, 8, 15, 24, 30}
// We get the above output using following
// arr[] = {2*3, 2*4, 3*5, 4*6, 5*6}
O(n) time and O(1) spacevoid modify(int arr[], int n){ // Nothing to do when array size is 1 if (n <= 1) return; // store current value of arr[0] and update it int prev = arr[0]; arr[0] = arr[0] * arr[1]; // Update rest of the array elements for (int i=1; i<n-1; i++) { // Store current value of next interation int curr = arr[i]; // Update current value using previos value arr[i] = prev * arr[i+1]; // Update previous value prev = curr; } // Update last array element arr[n-1] = prev * arr[n-1];}