Given three integers, find maximum and minimum number between them without using conditional statements or ternary operator.
Approach 1: (Using short-circuiting in Boolean expressions)
The idea is to take advantage of short-circuiting in Boolean expressions. We know that in Boolean AND operation such as x && y, y is evaluated only if x is true. If x is false, then y is not evaluated, because the whole expression would be false which can be deduced without even evaluating y. This is called short-circuiting in Boolean expressions.
The idea is to apply this principle to the below code. Initially max is a. Now if (max < b) is true, then that means, b is greater than a so the second sub-expression (max = b) is evaluated and max is set to b. If however (max < b) is false, then second sub-expression will not be evaluated and max will remain a (which is greater than b). In a similar fashion, the second expression is evaluated.
We can implement minimum function as well in similar fashion.
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#include <iostream> using namespace std; int maximum(int a, int b, int c) { // initialize max with a int max = a; // set max to b if and only if max is less than b (max < b) && (max = b); // these are not conditional statements. // set max to c if and only if max is less than c (max < c) && (max = c); // these are just Boolean expressions. return max; } int minimum(int a, int b, int c) { // initialize min with a int min = a; // set min to b if and only if min is more than b (min > b) && (min = b); // set min to c if and only if min is more than c (min > c) && (min = c); return min; } // main function int main() { cout << maximum(7, 9, 4) << endl; cout << minimum(6, 3, 9); return 0; } |
Approach 2: (Using array index)
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#include <iostream> using namespace std; int maximum(int a, int b, int c) { // arr1 will contain first two elements int arr1[] = { a, b }; // arr2 will contain maximum of first two elements at 0 index // and third element at index 1 int arr2[] = { arr1[a < b], c }; // finally return the maximum element return arr2[arr2[0] < c]; } int minimum(int a, int b, int c) { // arr1 will contain first two elements int arr1[] = { a, b }; // arr2 will contain minimum of first two elements at 0 index // and third element at index 1 int arr2[] = { arr1[a > b], c }; // finally return the minimum element return arr2[arr2[0] > c]; } // main function int main() { cout << maximum(6, 3, 9) << endl; cout << minimum(6, 3, 9); return 0; } |
Approach 3: (Approach 2 Simplified)
We can simplify approach 2 by breaking the problem into finding maximum/minimum of two numbers.
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#include <iostream> using namespace std; int maximum(int a, int b) { int lookup[] = {a, b}; return lookup[a < b]; } int maximum (int a, int b, int c) { return maximum(a, maximum(b, c)); } // main function int main() { cout << maximum(6, 3, 9) << endl; return 0; } |
We can implement minimum function as well in similar fashion.
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#include <iostream> using namespace std; int minimum(int a, int b) { int lookup[] = {a, b}; return lookup[a > b]; } int minimum(int a, int b, int c) { return minimum(a, minimum(b, c)); } // main function int main() { cout << minimum(6, 3, 9) << endl; return 0; } |
Approach 4: (Using repeated subtraction)
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#include <iostream> using namespace std; int minimum (int a, int b, int c) { int min = 0; while (a && b && c) a--, b--, c--, min++; return min; } int maximum (int a, int b, int c) { int max = 0; while (a > 0 || b > 0 || c > 0) a--, b--, c--, max++; return max; } // main function int main() { cout << maximum(6, 3, 9) << endl; cout << minimum(6, 3, 9); return 0; } |
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#include <iostream>using std::cout;
using std::endl;
template<typename T0, typename T1>
auto max(T0 a, T1 b) {
return 1./2. * (a + b + abs(a - b));
}
template<typename T0, typename T1, typename T2>
auto max3(T0 a, T1 b, T2 c) {
return max(a, max(b, c));
}
int main() {
cout << "max(2.001, 2): " << max(2.001, 2) << endl;
cout << "max(2, 1): " << max(2, 1) << endl;
cout << "max(2, 2): " << max(2, 2) << endl;
cout << "max(0, 0): " << max(0, 0) << endl;
cout << "max(-1, -10): " << max(-1, -10) << endl;
cout << "max(-1, 0): " << max(-1, 0) << endl;
cout << "max(1, -10): " << max(1, -10) << endl;
cout << "max3(2.001, 2, 4): " << max3(2.001, 2, 4) << endl;
cout << "max3(2, 1, 0): " << max3(2, 1, 0) << endl;
cout << "max3(2, 2, -1): " << max3(2, 2, -1) << endl;
cout << "max3(0, 0, 0): " << max3(0, 0, 0) << endl;
cout << "max3(-1, -10, -3): " << max3(-1, -10, -3) << endl;
cout << "max3(-1, 0, 10): " << max3(-1, 0, 10) << endl;
cout << "max3(1, -10, -2): " << max3(1, -10, -2) << endl;
}