Analyze the time complexity and space complexity of following algorithms. 3 1 /** Returns the sum of

Analyze the time complexity and space complexity of following algorithms.

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3 1 /** Returns the sum of the integers in given array. */ 2 public static int example1(int[] arr) { int n = arr.length, total = 0; for (int j=0; j<n; j++) total += arr[j]; return total; 4 5 6 7 8 9 /** Returns the sum of the integers with even index in given array. */ 10 public static int example2(int[] arr) { 11 12 13 14 15 } 16 17 /** Returns the sum of the prefix sums of given array. */ 18 public static int example3(int[] arr) { int n = arr.length, total = 0; for (int j=0; j<n; j++) int 19 for (int k=0; k <= j; k++) total += arr[j]; return total; } int n = arr.length, total = 0; for (int j=0; j<n; j+= 2) total += arr[j]; return total; 20 21 902 22:01 23 24 } 25 26 /** Returns the sum of the prefix sums of given array. */ 27 public static int example4(int[] arr) { 28 int n= arr.length, prefix = 0, total = 0; 29 for (int j=0; j <n; j++) { 30 prefix + = arr[j]; 31 total += prefix; 32 } return total; // loop from 0 to n-1 Sasod dos// loop from 0 to j Note ad nuo dur Sugidong s to ost al quis ono) algo heim () 20 // loop from 0 to n-1 total) count++; thoni gets Imantdia-s va // note the increment of 2 33 34 } 35 v dogo 36 /** Returns the number of times second array stores sum of prefix sums from first. */ 37 public static int example5(int[] first, int[] second) { // assume equal-length arrays 38 int n = first.length, count = 0; 39 for (int i 0; i < n; i++) { // loop from 0 to n-1 40 int total = 0; 41 for (int j=0; j <n; j++) 42 A 43 for (int k=0; k <= j; k++) total += first[k]; if (second[i] = 44 == 45 } 46 return count; 47 Scanned with CamScanner. // loop from 0 to n-1 // loop from 0 to n-1 // loop from 0 to j 101 if oding und Code Fragment 4.12: Some sample algorithms for analysis. 3 1 /** Returns the sum of the integers in given array. */ 2 public static int example1(int[] arr) { int n = arr.length, total = 0; for (int j=0; j<n; j++) total += arr[j]; return total; 4 5 6 7 8 9 /** Returns the sum of the integers with even index in given array. */ 10 public static int example2(int[] arr) { 11 12 13 14 15 } 16 17 /** Returns the sum of the prefix sums of given array. */ 18 public static int example3(int[] arr) { int n = arr.length, total = 0; for (int j=0; j<n; j++) int 19 for (int k=0; k <= j; k++) total += arr[j]; return total; } int n = arr.length, total = 0; for (int j=0; j<n; j+= 2) total += arr[j]; return total; 20 21 902 22:01 23 24 } 25 26 /** Returns the sum of the prefix sums of given array. */ 27 public static int example4(int[] arr) { 28 int n= arr.length, prefix = 0, total = 0; 29 for (int j=0; j <n; j++) { 30 prefix + = arr[j]; 31 total += prefix; 32 } return total; // loop from 0 to n-1 Sasod dos// loop from 0 to j Note ad nuo dur Sugidong s to ost al quis ono) algo heim () 20 // loop from 0 to n-1 total) count++; thoni gets Imantdia-s va // note the increment of 2 33 34 } 35 v dogo 36 /** Returns the number of times second array stores sum of prefix sums from first. */ 37 public static int example5(int[] first, int[] second) { // assume equal-length arrays 38 int n = first.length, count = 0; 39 for (int i 0; i < n; i++) { // loop from 0 to n-1 40 int total = 0; 41 for (int j=0; j <n; j++) 42 A 43 for (int k=0; k <= j; k++) total += first[k]; if (second[i] = 44 == 45 } 46 return count; 47 Scanned with CamScanner. // loop from 0 to n-1 // loop from 0 to n-1 // loop from 0 to j 101 if oding und Code Fragment 4.12: Some sample algorithms for analysis.

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