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You have given an array
Find the total pairs can be formed using this array and also print that all.
Print numbers of total pair exist in this array too.
Question 3: Java / C / C++ / Python / JavaScript.
**Note : For Python Treat this array as list then solve it.
{2, 3, 9, 6, 4, 5}
Find the total pairs can be formed using this array and also print that all.
Print numbers of total pair exist in this array too.
Question 3: Java / C / C++ / Python / JavaScript.
**Note : For Python Treat this array as list then solve it.
>> Write a program to print only next prime number. Example - input is 5 output is 7.
import math
def isPrime(n):
if n == 2:
return True
for i in range(2, int(math.sqrt(n))+1):
if n % i == 0:
return False
return True
def main():
num = int(input("Enter a number: "))
while True:
num += 1
if isPrime(num) == True:
print(num)
break
main()
Question 3 >> Solution!
import java.math.*;
public class TrappingRainwater {
public static int trappedRainwater(int arr[]) {
int trappedWater = 0;
int[] leftMax = new int[arr.length];
int[] rightMax = new int[arr.length];
// Auxiliary array
// left max
leftMax[0] = arr[0];
for(int i=1; i<arr.length; i++) {
leftMax[i] = Math.max(arr[i], leftMax[i-1]);
}
// right max
rightMax[arr.length-1] = arr[arr.length-1];
for(int i=arr.length-2; i>=0; i--) {
rightMax[i] = Math.max(rightMax[i+1], arr[i]);
}
// calculating rainwater
for(int i=0; i<arr.length; i++) {
int waterLevel = Math.min(leftMax[i], rightMax[i]);
trappedWater += Math.max(0, waterLevel-arr[i]);
}
return trappedWater;
}
public static void main(String[] arg) {
int[] array = {4, 2, 0, 6, 3, 2, 5};
int res = trappedRainwater(array);
System.out.print("Total trapped rainwater for\ngiven array is " + res);
}
}
Question 3 >> Solution!
# Calculate the rain water trapped by the gaps between the buildings
height = [4, 2, 0, 6, 3, 2, 5]
def trapping_rainwater(arr):
trapped_water = 0
right_max = [0] * len(arr)
left_max = [0] * len(arr)
# Auxiliary array
# left max boundary calculation.
left_max[0] = arr[0]
for i in range(1, len(arr)):
left_max[i] = max(arr[i], left_max[i-1])
# right max boundary calculation.
right_max[len(arr)-1] = arr[len(arr)-1]
for i in range(len(arr)-2, -1, -1):
right_max[i] = max(right_max[i+1], arr[i])
# Calculate the trapped water.
for i in range(len(arr)-1):
water_level = min(left_max[i], right_max[i])
trapped_water += max(0, water_level - arr[i])
print(trapped_water)
trapping_rainwater(height)
Question 4 : Buy & Sell Stocks
You are given an array prices where prices[i] is the price of a given stock
on the i'th day. You want to maximize your profit by choosing a single day to
buy one stock and choosing a different day in the future to sell that stock.
Return the maximum profit you can achieve from this transaction. If you
cannot achieve any profit, return 0.
Using : Java / C / C++ / Python / JavaScript
You are given an array prices where prices[i] is the price of a given stock
on the i'th day. You want to maximize your profit by choosing a single day to
buy one stock and choosing a different day in the future to sell that stock.
Return the maximum profit you can achieve from this transaction. If you
cannot achieve any profit, return 0.
price = {7, 1, 5, 3, 6, 4}
Using : Java / C / C++ / Python / JavaScript
Question 4 solution >>
public class Main {
public static void stocks(int arr[]) {
int maxProfit = 0;
int buyingPrice = Integer.MAX_VALUE;
// max profit
for(int i=0; i<arr.length; i++) {
if(buyingPrice < arr[i]) {
int profit = arr[i] - buyingPrice;
maxProfit = Math.max(maxProfit, profit);
} else {
buyingPrice = arr[i];
}
}
System.out.println("The Max Profit is " + maxProfit);
}
public static void main(String[] args) {
int arr[] = {7, 1, 5, 3, 6, 4};
stocks(arr);
}
}
Question 5 Solution »
public class DoubleElement {
public static boolean doubleElement(int[] arr) {
boolean res = false;
// outer loop
for(int i=0; i<arr.length; i++) {
for(int j=i+1; j<arr.length; j++) {
if(arr[i] == arr[j]) {
res = true;
break;
}
}
}
return res;
}
public static void main(String[] arg) {
int[] array = {1, 2, 3, 4};
int[] array2 = {1, 2, 3, 1};
System.out.println(doubleElement(array));
System.out.println(doubleElement(array2));
}
}
Question 5 Solution »
array = [1, 2, 3, 1]
array2 = [1, 2, 3, 4]
array3 = [1, 2, 3, 4, 5, 3]
def double_element(arr):
res = False
# outer loop
for i in range(len(arr)):
for j in range(i+1, len(arr)):
if arr[i] == arr[j]:
res = True
return res
print(double_element(array))
print(double_element(array2))
print(double_element(array3))
Question 6: Short an array in ascending order using bubble shorting algorithm.
Using : Java / C / C++ / Python / JavaScript
array = {5, 4, 3, 2, 1}
Using : Java / C / C++ / Python / JavaScript
Question 6 Solution »
// shorting an array using bubble shorting algorithm.
// array = {5, 4, 3, 2, 1}
#include <stdio.h>
void bubbleShort(int array[], int N) {
for(int i=0; i<N-1; i++) {
for(int j=i+1; j<N; j++) {
if(array[i] > array[j]) {
array[i] = array[i] + array[j];
array[j] = array[i] - array[j];
array[i] = array[i] - array[j];
}
}
}
}
int main() {
int array[5] = {5, 4, 3, 2, 1};
int N = 5;
bubbleShort(array, N);
for(int i=0; i<N; i++) {
printf("%d ", array[i]);
}
}
Question 6 Solution »
// Bubble shorting
let array = [5, 4, 1, 3, 2];
console.log(array);
for(let i=0; i<array.length; i++) {
for(let j=i+1; j<array.length; j++) {
if (array[i] > array[j]) {
array[i] = array[i] + array[j];
array[j] = array[i] - array[j];
array[i] = array[i] - array[j];
}
}
}
console.log(array);
Question 6 Solution »
# bubble short
array = [5, 4, 1, 3, 2]
def bubble_short(arr):
for i in range(len(arr)):
for j in range(i+1, len(arr)):
if arr[i] > arr[j]:
arr[i] = arr[i] + arr[j]
arr[j] = arr[i] - arr[j]
arr[i] = arr[i] - arr[j]
bubble_short(array)
print(array)
Question 6 Solution »
// bubble shorting
public class BubbleShort {
public static void bubbleShort(int[] arr) {
// outer loop
for(int i=0; i<arr.length; i++) {
// inner loop
for(int j=i+1; j<arr.length; j++) {
if(arr[i] > arr[j]) {
arr[i] = arr[i] + arr[j];
arr[j] = arr[i] - arr[j];
arr[i] = arr[i] - arr[j];
}
}
}
}
public static void main(String[] arg) {
int[] array = {5, 4, 1, 3, 2};
bubbleShort(array);
for(int i=0; i<array.length; i++) {
System.out.print(array[i] + " ");
}
}
}