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Copy file name to clipboardExpand all lines: LeetcodeProblems/Algorithms/easy/Best_Time_to_buy_and_sell_stock.js
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@@ -30,41 +30,44 @@ Constraints:
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*/
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/*
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Approach:
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let use initialize Left and Right pointer to first and second position of array
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Here Left is to buy stock and Right is to sell stock
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We will use a Two pointers strategy (Left and Right pointers).
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The first will start pointing to the first element, and the right to the second position of array.
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The Left is to buy stock and Right is to sell stock
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Then we initialize our maxProfitValue as 0.
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We initialize our maxProfitValue as 0.
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Now we will start our while loop, and we will run till our
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Right pointer less than the array's length.
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Now we will start our while loop and we will run till our
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Right pointer less then length of array
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For Example:
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prices=[7,1,5,3,6,4]
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Note:
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prices[left] --> buy stock
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prices[right] --> sell stock
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now we will check price at right and left pointer
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We will check the price at the right and left pointer
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step 1:
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price[left]=7 price[right]=1 profit=-6
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here price[left] is greater than price[right] so we will move left pointer to the right position and increment our right pointer by 1. We always want our left point to be minimum
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here, price[left] is greater than price[right], so we will move the left pointer to the right position
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and increment our right pointer by 1. We always want our left point to be the minimum.
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step 2:
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price[left]=1 price[right]=5 profit=4
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here price[left] is less than price[right] which means we will get profit so we will update our maxProfitValue and move our right pointer alone
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here, price[left] is less than price[right], which means we will get profit,
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so we will update our maxProfitValue and move our right pointer alone
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step 3:
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price[left]=1 price[right]=3 profit=2
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here price[left] is less than price[right] which means we will get profit so we will check our maxProfitValue previously it
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was 4 now our current profit is 2 so we will check which is maximum and update our maxProfitValue and move our right pointer alone
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here, price[left] is less than price[right], we will get profit, so we will compare the maxProfitValue with the current profit.
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We will update our maxProfitValue and move our right pointer alone
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step 4:
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price[left]=1 price[right]=6 profit=5
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here price[left] is less than price[right] which means we will get profit so we will check our maxProfitValue previously it was 4 now our current profit is 5 so we will check which is maximum and update our maxProfitValue and move our right pointer alone
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