Week 8 Prompts - Java

week-8/Java/Twumasi-Pennoh.java

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+package week8;
+import java.util.*;
+import java.lang.*;
+
+public class Week8 {
+
+	public static void main(String [] args) {
+//		int listSize = 20;
+//		int [] array = {1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4};
+//		int kTimes = 8;
+//		System.out.println(repeatedKTimes(listSize, array, kTimes));
+
+
+
+		//CASES FOR PROMPT 1
+		/*
+		 * first case
+		 */
+//		int [] daysAndQueries = {5, 4};
+//		int [] numPerDay = {1, 2, 3, 4, 5};
+//		int [] stonesCollected = {3, 8, 10, 14};
+//		stonesLove(daysAndQueries, numPerDay, stonesCollected);
+
+
+		/*
+		 * second case
+		 */
+//		int [] daysAndQueries = {20, 16};
+//		int [] numPerDay = {40, 31, 39, 7, 27, 10, 15, 22, 2, 25, 41, 44, 29, 32, 45, 10, 38, 36, 6, 1};
+//		int [] stonesCollected = {20, 418, 439, 251, 363, 331, 101, 328, 49, 460, 244, 390, 142, 384, 47, 470, 295, 263, 343, 97};
+//		stonesLove(daysAndQueries, numPerDay, stonesCollected);
+
+	}
+
+
+	/*
+	 * Promp1: Stone's Love
+	 * You are given the number of stones, she collects each day for N numbers of days, 
+	 * starting from the very first day. Now you are given Q queries, in each query her 
+	 * friend shambhavi asks you the number of days she has taken to collect M number of 
+	 * stones. Please note that each query contains the different number of M.
+	 * 
+	 * Input: 
+	 * First line contains N and Q, the number of days and number of queries. Second line 
+	 * contains N integers, in which ith integer denotes the number of stones she has 
+	 * collected on ith day. Then next line contains Q space-separated integers, M, where 
+	 * ith M denotes the ith query, i.e., number of stones.
+	 * 
+	 * Output:
+	 * For each of the Q queries, you have to output the number of days she has taken to 
+	 * collect M number of stones in a new line.
+	 */
+
+	public static int numDaysToCollectStones(int[] numPerDays, int[] orig, int numStones){
+        int maxIndex = numPerDays.length-1;
+        int minIndex = 0;
+        int returnNum = -1;
+        while(maxIndex >= minIndex){
+            int middleIndex = (maxIndex + minIndex)/2;
+            boolean firstPassCase = numPerDays[middleIndex] == numStones;
+            boolean secondPassCaseA = (numPerDays[middleIndex] - numStones) >= 0; 
+            boolean secondPassCaseB = middleIndex != 0 && numPerDays[middleIndex-1] != numStones && (numPerDays[middleIndex-1]+orig[middleIndex] >= numStones);
+            if(secondPassCaseA){
+            	if(numPerDays[middleIndex] == numStones) {
+            		returnNum = middleIndex+1;
+            		break;
+            	}else {
+            		returnNum = middleIndex+1;
+            		maxIndex = middleIndex-1;
+            	}
+            }
+            else if(numPerDays[middleIndex] > numStones){
+                maxIndex = middleIndex-1;
+            }
+            else if(numPerDays[middleIndex] < numStones){
+                minIndex = middleIndex+1;
+            }
+        }
+        return returnNum;
+    }
+
+	public static void stonesLove(int[] daysAndQueries, int[] numPerDay, int[] stonesCollected) {
+		int[] orig = new int[numPerDay.length];
+        for(int i = 0; i < numPerDay.length; i++){
+            orig[i] = numPerDay[i];
+            if(i != 0){
+                numPerDay[i] = numPerDay[i-1] + numPerDay[i];
+            }
+        }
+
+        for(int i = 0; i < daysAndQueries[1]; i++){
+            System.out.println(numDaysToCollectStones(numPerDay, orig, stonesCollected[i]));
+        }
+	}
+
+
+
+	/*
+	 * Prompt 2: Repeated K Times
+	 * Given a List of N number a1,a2,a3........an, You have to find smallest number from the
+	 * List that is repeated in the List exactly K number of times.
+	 */
+	public static int repeatedKTimes(int listSize, int[] integerArray, int kTimes) {
+		Arrays.sort(integerArray);
+        int currentInt = 0;
+        int currentIntCount = 0;
+        for(int i = 0; i < integerArray.length; i++){
+            if(currentInt != integerArray[i] && currentIntCount == kTimes){
+                break;
+            }
+            else if(currentInt != integerArray[i] && currentIntCount != kTimes){
+                currentInt = integerArray[i];
+                currentIntCount = 1;
+            } 
+            else{
+                currentIntCount++;
+            }
+        }
+
+        return currentInt;
+	}
+}