respectively. boolean arraysEqual(int[] array, int[] array2) { int n = array1.length; int m = array2.length; if(n...

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respectively. boolean arraysEqual(int[] array¹, int[] array2) { int n = array1.length; int m = array2.length; if(n != m) return false; Consider the following algorithm, where the input is two arrays of sizes n and m, for(int k = 0; k < n; k++){ if(array1[k] != array2[k]) return false; } return true; } 1. (5 points) What is the best case time complexity of this algorithm? Justify your answer. 2. (5 points) What is the worst case time complexity of this algorithm? Justify your answer. 3. (5 points) What is the space complexity of this algorithm? Justify your answer. respectively. boolean arraysEqual(int[] array¹, int[] array2) { int n = array1.length; int m = array2.length; if(n != m) return false; Consider the following algorithm, where the input is two arrays of sizes n and m, for(int k = 0; k < n; k++){ if(array1[k] != array2[k]) return false; } return true; } 1. (5 points) What is the best case time complexity of this algorithm? Justify your answer. 2. (5 points) What is the worst case time complexity of this algorithm? Justify your answer. 3. (5 points) What is the space complexity of this algorithm? Justify your answer.

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1 BestCase Scenario Time Complexity When the first member of both arrays differs ... View the full answer