- In the exercises below there is given a code for methods named find1, find2, find3 and printMany. Analyze each of the corresponding algorithms according to the points as follows:

1. Choose the input size n and explain your choice. Estimate the running time (number of steps) T(n) in terms of the O(n) scale. Use the simplest and possibly the smallest valid big-Oh expressions.

2. If it applies, point out your estimates both for the worst case and best case.

3. Document and comment each method. Describe the tasks of the methods, explain the meaning the return value if applies, show and justify your big-Oh estimate.

4. It is not necessary to run these methods in actual programs, but if the task it performs is dubious, testing the method with various input in actual applications of the code may help to find its purpose and the big-Oh estimate.

Exercises

1.

int find1( int[] list, int element ){

int answer = 0;

for(int k = 0; k < list.length; k++ ) if (element == list[k]) answer++; return answer; }//end method

Comments:

2.

public int find2(int[] arr){

int zeroCounter = find1(arr, 0);

if (zeroCounter > arr.length - 2)

return 0;

while(zeroCounter < arr.length - 2){

//see maxIndex() definition below

int max = maxIndex(arr);

arr[max] = 0;

//see display() definition below

display(arr);

zeroCounter ++;

}

return maxIndex(arr);

}//end method

//helper methods int maxIndex(int[]arr){ int maxindex = 0;

for(int k = 0 ; k< arr.length; k++){ // note the use of absolute value if(Math.abs(arr[maxindex]) < Math.abs(arr[k]))

maxindex = k;

}

return maxindex;

}

void display(int[]arr){ System.out.println(); for(int k = 0 ; k< arr.length; k++) System.out.print(arr[k]+ );

System.out.println();

}

Comments

3.

int find(int[] nums){

int answer = nums[0];

int temp = 0;

for(int k = 0; k < nums.length; k++ ) for(int j = k; j< nums.length; j++){

//see helper method subSum below

temp = subSum(nums, k, j ); if (temp > answer)

answer = temp;

} return answer;

}

Note: Given two indices i<=j of an array of integers num, the sum num[i]+ num[i+1] + + num[j] is called a sub-sum

//helper method int subSum(int[]arr, int i, int j){

int sum = 0; for(int k = i; k<= j; k++) sum += arr[k];

return sum;

}

Comments

4.

void printMany(int[]arr){

int N = arr.length; for(int k = 0 ; k< N; k++){

int p = k;

while(p>0){

System.out.println(arr[p]+" ");

p = p/2;

}

}

}

Comments