Webif(n==1) return 0; int o1=countStepsTo1(n-1); int minSteps=o1; if(n%3==0){int o3=countStepsTo1(n/3); if(minSteps>o3) minSteps=o3;} if(n%2==0){int … WebIf we start with 0, valid numbers will be 00, 08 (count: 2) If we start with 1, valid numbers will be 11, 12, 14 (count: 3) If we start with 2, valid numbers will be 22, 21, 23,25 (count: 4) If we start with 3, valid numbers will be 33, 32, 36 (count: 3) If we start with 4, valid numbers will be 44,41,45,47 (count: 4) If we start with 5, valid …
Minimum step to reach one - GeeksforGeeks
WebMay 12, 2024 · Minimum steps to minimize n as per given condition. Given a number n, count minimum steps to minimize it to 1 according to the following criteria: If n is … Webstatic int minSteps (int n) { if (n == 1) { return 0; } int minus = 1 + minSteps(n - 1); int divideBy3 = Integer.MAX_VALUE; if (n % 3 == 0) { divideBy3 = 1 + minSteps(n / 3); } int … hidamari ga kikoeru cap 1
Competitive-Coding/staircase.cpp at master - GitHub
WebAnalyze the problem and see the order in which the sub-problems are solved and start solving from the trivial subproblem, up towards the given problem. In this process, it is … WebJul 19, 2024 · Output: 1 Explanation: Step 1: 0 + 1 = 1 = K Input: K = 4 Output: 3 Explanation: Step 1: 0 + 1 = 1, Step 2: 1 * 2 = 2, Step 3: 2 * 2 = 4 = K Recommended: Please try your approach on {IDE} first, before moving on to the solution. Approach: If K is an odd number, the last step must be adding 1 to it. Weba = countStepsTo1(n-1); if(n%2==0) b=countStepsTo1(n/2); if(n%3==0) c=countStepsTo1(n/3); return 1+min(a,min(b,c));} //using dynamic programming: … hidamari cooking