Given an array of non-negative integers. Find the largest multiple of 3 that can be formed from array elements.
For example, if the input array is {8, 1, 9}, the output should be "9 8 1″, and if the input array is {8, 1, 7, 6, 0}, output should be "8 7 6 0″.
For example, if the input array is {8, 1, 9}, the output should be "9 8 1″, and if the input array is {8, 1, 7, 6, 0}, output should be "8 7 6 0″.
1) A number is multiple of 3 if and only if the sum of digits of number is multiple of 3.
2) If a number is multiple of 3, then all permutations of it are also multiple of 3.
3) We get the same remainder when we divide the number and sum of digits of the number. For example, if divide number 151 and sum of it digits 7, by 3, we get the same remainder 1.
What is the idea behind above facts?
The value of 10%3 and 100%3 is 1. The same is true for all the higher powers of 10, because 3 divides 9, 99, 999, … etc.
Let us consider a 3 digit number n to prove above facts. Let the first, second and third digits of n be ‘a’, ‘b’ and ‘c’ respectively. n can be written as
The value of 10%3 and 100%3 is 1. The same is true for all the higher powers of 10, because 3 divides 9, 99, 999, … etc.
Let us consider a 3 digit number n to prove above facts. Let the first, second and third digits of n be ‘a’, ‘b’ and ‘c’ respectively. n can be written as
n = 100.a + 10.b + c
Since (10^x)%3 is 1 for any x, the above expression gives the same remainder as following expression
1.a + 1.b + c
So the remainder obtained by sum of digits and ‘n’ is same.
Following is a solution based on the above observation.
1. Sort the array in non-decreasing order.
2. Take three queues. One for storing elements which on dividing by 3 gives remainder as 0.The second queue stores digits which on dividing by 3 gives remainder as 1. The third queue stores digits which on dividing by 3 gives remainder as 2. Call them as queue0, queue1 and queue2
3. Find the sum of all the digits.
4. Three cases arise:
……4.1 The sum of digits is divisible by 3. Dequeue all the digits from the three queues. Sort them in non-increasing order. Output the array.
……4.1 The sum of digits is divisible by 3. Dequeue all the digits from the three queues. Sort them in non-increasing order. Output the array.
……4.2 The sum of digits produces remainder 1 when divided by 3.
Remove one item from queue1. If queue1 is empty, remove two items from queue2. If queue2 contains less than two items, the number is not possible.
Remove one item from queue1. If queue1 is empty, remove two items from queue2. If queue2 contains less than two items, the number is not possible.
……4.3 The sum of digits produces remainder 2 when divided by 3.
Remove one item from queue2. If queue2 is empty, remove two items from queue1. If queue1 contains less than two items, the number is not possible.
Remove one item from queue2. If queue2 is empty, remove two items from queue1. If queue1 contains less than two items, the number is not possible.
5. Finally empty all the queues into an auxiliary array. Sort the auxiliary array in non-increasing order. Output the auxiliary array.
typedef struct Queue{ int front; int rear; int capacity; int* array;} Queue;// This function puts all elements of 3 queues in the auxiliary arrayvoid populateAux (int* aux, Queue* queue0, Queue* queue1, Queue* queue2, int* top ){ // Put all items of first queue in aux[] while ( !isEmpty(queue0) ) aux[ (*top)++ ] = Dequeue( queue0 ); // Put all items of second queue in aux[] while ( !isEmpty(queue1) ) aux[ (*top)++ ] = Dequeue( queue1 ); // Put all items of third queue in aux[] while ( !isEmpty(queue2) ) aux[ (*top)++ ] = Dequeue( queue2 );}// The main function that finds the largest possible multiple of// 3 that can be formed by arr[] elementsint findMaxMultupleOf3( int* arr, int size ){ // Step 1: sort the array in non-decreasing order qsort( arr, size, sizeof( int ), compareAsc ); // Create 3 queues to store numbers with remainder 0, 1 // and 2 respectively Queue* queue0 = createQueue( size ); Queue* queue1 = createQueue( size ); Queue* queue2 = createQueue( size ); // Step 2 and 3 get the sum of numbers and place them in // corresponding queues int i, sum; for ( i = 0, sum = 0; i < size; ++i ) { sum += arr[i]; if ( (arr[i] % 3) == 0 ) Enqueue( queue0, arr[i] ); else if ( (arr[i] % 3) == 1 ) Enqueue( queue1, arr[i] ); else Enqueue( queue2, arr[i] ); } // Step 4.2: The sum produces remainder 1 if ( (sum % 3) == 1 ) { // either remove one item from queue1 if ( !isEmpty( queue1 ) ) Dequeue( queue1 ); // or remove two items from queue2 else { if ( !isEmpty( queue2 ) ) Dequeue( queue2 ); else return 0; if ( !isEmpty( queue2 ) ) Dequeue( queue2 ); else return 0; } } // Step 4.3: The sum produces remainder 2 else if ((sum % 3) == 2) { // either remove one item from queue2 if ( !isEmpty( queue2 ) ) Dequeue( queue2 ); // or remove two items from queue1 else { if ( !isEmpty( queue1 ) ) Dequeue( queue1 ); else return 0; if ( !isEmpty( queue1 ) ) Dequeue( queue1 ); else return 0; } } int aux[size], top = 0; // Empty all the queues into an auxiliary array. populateAux (aux, queue0, queue1, queue2, &top); // sort the array in non-increasing order qsort (aux, top, sizeof( int ), compareDesc); // print the result printArr (aux, top); return 1;}
2) We can avoid extra space for queues. We know at most two items will be removed from the input array. So we can keep track of two items in two variables.
3) At the end, instead of sorting the array again in descending order, we can print the ascending sorted array in reverse order. While printing in reverse order, we can skip the two elements to be removed.
The above code works only if the input arrays has numbers from 0 to 9. It can be easily extended for any positive integer array. We just have to modify the part where we sort the array in decreasing order, at the end of code.
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