1. Analyze the running time of insertion sort on an array of...

1. Analyze the running time of insertion sort on an array of size n. Show detailed analysis of the steps and express the final answer using order notation. Assume all basic operations take unit time (c = 1). void insertionSort(int[] arr) { // Sort arr[0..arr.length-1]. for(int i=1; i<arr.length; i++) { int key = arr[i]; int j = i-1; // Insert key into the right spot while (j >= 0 && arr[j] > key) { arr[j+1] = arr[j]; j = j-1; } arr[j+1] = key; } }

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00:01 Hello students to analyze the running time of the insertion short on the array of the size n so we need to break down the algorithm as and count the basic operation so as the first is the outer loop the outer loop runs for i equal to 1 to i equal to n minus 1 so it reads through each element of an array once so inside the outer loop where you can see the constants of the operations like the assigning of key equal to array of i and assignment of j is equal to j minus 1 so when so these have been inside so in the inner loop runs inner loop runs in a worst -case scenario when the array is sorted in a descending order so in the case it performs the operation as the comparison of the j is lesser it is greater than or equal to 0 which is a single comparison and the comparison of array of array of a j is greater than a key which is another single comparison so if the condition in in the inner loop are met we perform the operation as the assignment of the assignment of the array of a j plus 1 equal to the array of j so a decrement whereas the decrement of the j equal to it is equal to j minus 1 finally after executing the inner loop we perform the assignment of the array of the j plus 1 with a key so whereas it has the worst -case time complexity of an insertion so as the outer loop runs for the n minus 1 times from i equal to 1 to i is equal to n minus 1 so the worst -case scenario the inner loop may have the traverse the entire entire sorted portion of the array.

02:03 So this means that it executes up to the i comparison so therefore the total number of the comparison made in an inner loop over all the iteration of the outer loop is the sum of the integration from 1 to n n minus 1 so which is which is from 1 to 1 to n minus 1 so this summation this sum is equivalent to the it is equivalent to n minus 1 into n by 2 so which has an which has an time complexity of big o of n to the power of 2 so it has the time complexity of the big o of n to the power of 2 fine so it has the worst -case complexity of an insertion...

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