Answer:
I guess if there is experiment going on in absence one of those furniture then the experiment isn't successful
Answer:
8/9
Explanation:
i remember this question on my daughters book
Answer:
The solution code is written in Java
- public static void checkCommonValues(int arr1[], int arr2[]){
- if(arr1.length < arr2.length){
- for(int i = 0; i < arr1.length; i++){
- for(int j = 0; j < arr2.length; j++){
- if(arr1[i] == arr2[j]){
- System.out.print(arr1[i] + " ");
- }
- }
- }
- }
- else{
- for(int i = 0; i < arr2.length; i++){
- for(int j = 0; j < arr1.length; j++){
- if(arr2[i] == arr1[j]){
- System.out.print(arr2[i] + " ");
- }
- }
- }
- }
- }
Explanation:
The key idea of this method is to repeated get a value from the shorter array to check against the all the values from a longer array. If any comparison result in True, the program shall display the integer.
Based on this idea, an if-else condition is defined (Line 2). Outer loop will traverse through the shorter array (Line 3, 12) and the inner loop will traverse the longer array (Line 4, 13). Within the inner loop, there is another if condition to check if the current value is equal to any value in the longer array, if so, print the common value (Line 5-7, 14-16).
Answer: number 2 is the correct way to do it
Explanation:
