Find the Largest Number in a Rotated List - Amazon Top Interview Questions

Problem Statement :

You are given a list of unique integers nums that is sorted in ascending order and is rotated at some pivot point. Find the maximum number in the rotated list.

Can you solve it in \mathcal{O}(\log{}n)O(logn)?


n ≤ 100,000 where n is the length of nums.

Example 1


arr = [6, 7, 8, 1, 4]




The original sorted array of [1, 4, 6, 7, 8] was rotated at index 2 and results in the input array [6, 7, 8, 1, 4,]. And the largest number is 8.

Example 2


arr = [1, 2, 3]



Example 3


arr = [1]



Example 4


arr = [10, 1, 2, 3, 4, 7]



Solution :


                        Solution in C++ :

int solve(vector<int>& a) {
    int low = 0;
    int high = a.size() - 1;
    int maxi = -1000;
    while (high >= low) {
        int mid = (low + high) / 2;
        if (a[mid] < a[0]) {
            high = mid - 1;

        else if (a[mid] >= a[0]) {
            maxi = max(maxi, a[mid]);
            low = mid + 1;

    return maxi;

                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] arr) {
        int n = arr.length;
        int low = 0;
        int high = arr.length - 1;

        while (low < high) {
            int mid = (low + high) / 2;

            if (arr[mid] > arr[high]) {
                low = mid + 1;
            } else
                high = mid;
        if (low == 0)
            return arr[n - 1];
        return arr[low - 1];

                        Solution in Python : 
class Solution:
    def solve(self, arr):
        lo, hi = 0, len(arr)
        left, right = arr[0], arr[-1]

        while lo < hi:
            mid = (lo + hi) // 2
            if left > arr[mid]:
                hi = mid
                lo = mid + 1

        return arr[lo - 1]

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