Uber Interview Question for Software Developers


Country: India
Interview Type: Phone Interview




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0
of 0 vote

package main

import "fmt"

// Given an array of integers find the maximum value of
// a[i] - a[j] + a[k]
// with constraints
// i < j < k and a[i] < a[j] < a[k]
func main() {
	a := []int{1, 3, 4, 5, 4}

	max := 0 // Will return 0 if the matching combination is never found
	for i := 0; i < len(a)-2; i++ {
		for j := i + 1; j < len(a)-1; j++ {
			for k := j + 1; k < len(a); k++ {
				if a[i] < a[j] && a[j] < a[k] && max < a[i]-a[j]+a[k] {
					max = a[i] - a[j] + a[k]
				}
			}
		}
	}
	// O(n^3)
	fmt.Printf("Max is %d\n", max)

	// Optimization
	lookup := make([]int, len(a))
	m := a[len(a)-1]
	lookup[len(a)-1] = m
	for i := len(a) - 2; i > 0; i-- {
		if m < a[i] {
			m = a[i]
		}
		lookup[i] = m
	}
	fmt.Printf("Lookip is %v\n", lookup)
	max = 0
	for i := 0; i < len(a)-2; i++ {
		for j := i + 1; j < len(a)-1; j++ {
			if a[i] < a[j] && a[j] < lookup[j+1] && max < a[i]-a[j]+lookup[j+1] {
				max = a[i] - a[j] + lookup[j+1]
			}
		}
	}
	// O(n^2) + S(n)
	fmt.Printf("Max is %d\n", max)
}

- Ilya August 02, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

package main

import "fmt"

// Given an array of integers find the maximum value of
// a[i] - a[j] + a[k]
// with constraints
// i < j < k and a[i] < a[j] < a[k]
func main() {
	a := []int{1, 3, 4, 5, 4}

	max := 0 // Will return 0 if the matching combination is never found
	for i := 0; i < len(a)-2; i++ {
		for j := i + 1; j < len(a)-1; j++ {
			for k := j + 1; k < len(a); k++ {
				if a[i] < a[j] && a[j] < a[k] && max < a[i]-a[j]+a[k] {
					max = a[i] - a[j] + a[k]
				}
			}
		}
	}
	// O(n^3)
	fmt.Printf("Max is %d\n", max)

	// Optimization
	lookup := make([]int, len(a))
	m := a[len(a)-1]
	lookup[len(a)-1] = m
	for i := len(a) - 2; i > 0; i-- {
		if m < a[i] {
			m = a[i]
		}
		lookup[i] = m
	}
	fmt.Printf("Lookip is %v\n", lookup)
	max = 0
	for i := 0; i < len(a)-2; i++ {
		for j := i + 1; j < len(a)-1; j++ {
			if a[i] < a[j] && a[j] < lookup[j+1] && max < a[i]-a[j]+lookup[j+1] {
				max = a[i] - a[j] + lookup[j+1]
			}
		}
	}
	// O(n^2) + S(n)
	fmt.Printf("Max is %d\n", max)
}

- ilya.pimenov@gmail.com August 02, 2019 | Flag Reply
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0
of 0 vote

/**
     * @author Omid Ghiasi Tarzi
     */
    static int calculate(int[] arr) {
        int max = 0;
        boolean isMaxEmpty = true;
        for (int i = 0; i < arr.length - 2; i++) {
            for (int j = i + 1; j < arr.length - 1; j++) {
                for (int k = j + 1; k < arr.length; k++) {
                    if (arr[i] < arr[j] && arr[j] < arr[k]) {
                        int value = arr[i] - arr[j] + arr[k];
                        max = isMaxEmpty || value > max ? value : max;
                        isMaxEmpty = false;
                    }
                }
            }
        }
        return max;
    }

- omid095 October 12, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static int maxSum(int arr[]) {
		int max = Integer.MIN_VALUE;
		int left[] = new int[arr.length];
		int right[] = new int[arr.length];
		left[0] = arr[0];
		for (int i = 1; i < arr.length; i++) {
			left[i] = Math.max(arr[i], left[i-1]);
		}
		right[arr.length-1] = arr[arr.length-1];
		for (int i = arr.length-2; i>=0; i--) {
			right[i] = Math.max(arr[i], right[i+1]);
		}
		
		for (int i = 1; i < arr.length-1; i++) {
			int sum = left[i-1] -arr[i] + right[i+1];
			max = Math.max(max, sum);
		}
		
		return max;

- Anonymous February 17, 2020 | Flag Reply


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