The language I\'m using is C++. Implement a function bool areEqual(int nums1[],

Question

The language I'm using is C++.

Implement a function

bool areEqual(int nums1[], int length1, int nums2[], int length2);

that returns true if the two arrays have the same length and the same sequence of values. Otherwise, return false.

Explanation / Answer

// Given an array of ints, return true if 6 appears as either the first or last element in the array. // The array will be length 1 or more. public boolean firstLast6(int[] nums) { return (nums[0] == 6 || nums[nums.length-1] == 6); } // Given an array of ints, return true if the array is length 1 or more, and the // first element and the last element are the same. public boolean sameFirstLast(int[] nums) { return (nums.length >= 1 && nums[0] == nums[nums.length-1]);} // Return an int array length 3 containing the first 3 digits of pi, {3, 1, 4}. public int[] makePi() { int[] pi = {3, 1, 4}; return pi; } // Given 2 arrays of ints, a and b, return true if they have the same first element or they have the same last element. // Both arrays will be length 1 or more. public boolean commonEnd(int[] a, int[] b) { return (a[0] == b[0] || a[a.length-1] == b[b.length-1]); } // Given an array of ints length 3, return the sum of all the elements. public int sum3(int[] nums) { return (nums[0] + nums[1] + nums[2]); } // Given an array of ints length 3, return an array with the elements // "rotated left" so {1, 2, 3} yields {2, 3, 1}. public int[] rotateLeft3(int[] nums) { int[] rotated = {nums[1], nums[2], nums[0]}; return rotated; } // Given an array of ints length 3, return a new array with the elements in reverse order, // so {1, 2, 3} becomes {3, 2, 1}. public int[] reverse3(int[] nums) { int[] reversed = {nums[2], nums[1], nums[0]}; return reversed; } // Given an array of ints length 3, figure out which is larger between the first and last elements // in the array, and set all the other elements to be that value. Return the changed array. public int[] maxEnd3(int[] nums) { int[] maxVal = new int[3]; maxVal[0] = nums[0]; if(nums[2] >= maxVal[0]) maxVal[0] = nums[2]; maxVal[1] = maxVal[0]; maxVal[2] = maxVal[0]; return maxVal; } // Given an array of ints, return the sum of the first 2 elements in the array. // If the array length is less than 2, just sum up the elements that exist, // returning 0 if the array is length 0. public int sum2(int[] nums) { if(nums.length >= 2) return (nums[0] + nums[1]); if(nums.length == 1) return nums[0]; return 0; } // Given 2 int arrays, a and b, each length 3, return a new array length 2 // containing their middle elements. public int[] middleWay(int[] a, int[] b) { int[] mids = {a[1], b[1]}; return mids; } // Given an array of ints, return a new array length 2 containing the first and last // elements from the original array. The original array will be length 1 or more. public int[] makeEnds(int[] nums) { int[] temp = {nums[0], nums[nums.length-1]}; return temp; } // Given an int array length 2, return true if it contains a 2 or a 3. public boolean has23(int[] nums) { if(nums[0] == 2 || nums[0] == 3) return true; return (nums[1] == 2 || nums[1] == 3); } // Given an int array length 2, return true if it does not contain a 2 or 3. public boolean no23(int[] nums) { if(nums[0] == 2 || nums[0] == 3) return false; return !(nums[1] == 2 || nums[1] == 3); } // Given an int array, return a new array with double the length where its last element is the same as // the original array, and all the other elements are 0. The original array will be length 1 or more. // Note: by default, a new int array contains all 0's. public int[] makeLast(int[] nums) { int len = nums.length*2; int[] dubsArr = new int[len]; dubsArr[len-1] = nums[nums.length-1]; return dubsArr; } // Given an int array, return true if the array contains 2 twice, or 3 twice. // The array will be length 0, 1, or 2. public boolean double23(int[] nums) { if(nums.length == 2) { if(nums[0] == 2 && nums[1] == 2) return true; return (nums[0] == 3 && nums[1] == 3); } return false; } // Given an int array length 3, if there is a 2 in the array immediately followed by a 3, // set the 3 element to 0. Return the changed array. public int[] fix23(int[] nums) { int[] fxArr = {nums[0], nums[1], nums[2]}; if(nums[0] == 2 && nums[1] == 3) fxArr[1] = 0; if(nums[1] == 2 && nums[2] == 3) fxArr[2] = 0; return fxArr; } // Start with 2 int arrays, a and b, of any length. // Return how many of the arrays have 1 as their first element. public int start1(int[] a, int[] b) { int if(a.length >= 1 && a[0] == 1) ones += 1; if(b.length >= 1 && b[0] == 1) ones += 1; return ones; } // Start with 2 int arrays, a and b, each length 2. // Consider the sum of the values in each array. Return the array which has the largest sum. // In event of a tie, return a. public int[] biggerTwo(int[] a, int[] b) { int sum = a[0]+a[1]-b[0]-b[1]; if(sum >= 0) return a; return b; } // Given an array of ints of even length, return a new array length 2 containing the middle // two elements from the original array. The original array will be length 2 or more. public int[] makeMiddle(int[] nums) { int[] midArr = new int[2]; int half = nums.length/2; midArr[0] = nums[half-1]; midArr[1] = nums[half]; return midArr; } // Given 2 int arrays, each length 2, return a new array length 4 containing all their elements. public int[] plusTwo(int[] a, int[] b) { int[] combArr = {a[0], a[1], b[0], b[1]}; return combArr; } // Given an array of ints, swap the first and last elements in the array. // Return the modified array. The array length will be at least 1. public int[] swapEnds(int[] nums) { int temp = nums[0]; nums[0] = nums[nums.length-1]; nums[nums.length-1] = temp; return nums; } //Given an array of ints of odd length, return a new array length 3 containing the elements from // the middle of the array. The array length will be at least 3. public int[] midThree(int[] nums) { int[] halfArr = new int[3]; int half = nums.length/2; halfArr[0] = nums[half-1]; halfArr[1] = nums[half]; halfArr[2] = nums[half+1]; return halfArr; } // Given an array of ints of odd length, look at the first, last, and middle values in // the array and return the largest. The array length will be a least 1. public int maxTriple(int[] nums) { int max = nums[0]; if(max

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