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    <p>One more efficient implementation </p> <pre><code>int i, j; /* new length of modified array */ int NewLength = 1; for(i=1; i&lt; Length; i++){ for(j=0; j&lt; NewLength ; j++) { if(array[i] == array[j]) break; } /* if none of the values in index[0..j] of array is not same as array[i], then copy the current value to corresponding new position in array */ if (j==NewLength ) array[NewLength++] = array[i]; } </code></pre> <p>In this implementation there is no need for sorting the array. Also if a duplicate element is found, there is no need for shifting all elements after this by one position.</p> <p>The output of this code is array[] with size NewLength</p> <p>Here we are starting from the 2nd elemt in array and comparing it with all the elements in array up to this array. We are holding an extra index variable 'NewLength' for modifying the input array. NewLength variabel is initialized to 0.</p> <p>Element in array[1] will be compared with array[0]. If they are different, then value in array[NewLength] will be modified with array[1] and increment NewLength. If they are same, NewLength will not be modified.</p> <p>So if we have an array [1 2 1 3 1], then</p> <p>In First pass of 'j' loop, array[1] (2) will be compared with array0, then 2 will be written to array[NewLength] = array[1] so array will be [1 2] since NewLength = 2</p> <p>In second pass of 'j' loop, array[2] (1) will be compared with array0 and array1. Here since array[2] (1) and array0 are same loop will break here. so array will be [1 2] since NewLength = 2</p> <p>and so on</p>
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    1. POAlgorithm: efficient way to remove duplicate integers from an array
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    1. USByju
    1. USByju
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    1. CONice one. I have a suggestion to improve. Second nested loop can be changed to for(j=0; j < NewLength; j++) and last if checking can be changed to if (j == NewLength )
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      1. USVadakkumpadath
    2. COThat was a great suggession. I have updated the code based on ur comment
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      1. USByju
    3. COFail at least if we have same values in array {1,1,1,1,1,1}. Useless code.
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      1. USYuriy Chernyshov
 

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