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    <p>The algorithm can be summarrized in pseudo-code as :</p> <ol> <li>mark the current position</li> <li>go backward one character at a time until a space is found or end of input is reached</li> <li>now go forward copying each character to output, then go back to 1., except if eoi is reached</li> </ol> <p>So the input is traversed once in reverse, and one more time forward, but without coming back to a previously read position in either step 2. or 3. And when switching from step 3. to 1. it directly adjust the iterator. The <code>count</code> variable is used to track the state of the algorithm (it is in fact a simple state-machine). It is also reused to define the direction of the iteration.</p> <p>So, the algorithm is in fact <code>O(n)</code>.</p> <hr> <p>For more clarity, it could be rewritten as this, without changing the complexity:</p> <pre><code>void printStringWithWordReversed(const char* a) { int i,j,display_ToVal= strlen(a)-1, display_FromVal; for( i=display_ToVal; i&gt;=0 ; i=i+-1) { if ( (a[i] == ' ' || i == 0)) { // When entering this branch, we are switching from state 2 to // state 3 (this is the content of the first branch). cout &lt;&lt; " "; display_FromVal = i; if ( i == 0 ) cout &lt;&lt; a[i]; // This loop correspond to the state 3, and is equivalent to the // previous code in the particular case when count == 1. for (j = display_FromVal+1; j &lt;= display_ToVal; j=j+1) { cout &lt;&lt; a[j]; } // This postlude correspond to the transition from state 3 to state 1 // and correspond to the second branch in the original algorithm. display_ToVal = display_FromVal - 1; if ( i == 0 ) break; continue; } } } </code></pre> <p>So we lookup each word starting from the end and output them in correct order. This is clearly <code>O(n)</code> with both implementation (both in time and space if we assume that <code>cout</code> insertion operator for <code>char</code> is <code>O(1)</code>) since adding a <em>fixed number</em> (here two) of <code>O(n)</code> algorithm is still <code>O(n)</code> (the constant is ignored).</p>
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    1. POTime complexity of the code below?
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    1. USSylvain Defresne
    1. USSylvain Defresne
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      1. VTUpMod
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    1. COfirst solution that notices i doesn't change by a constant value. +1
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      1. USamit
    2. COFirst solution that *explains* and doesn't jump into the "one loop = O(N)" fallacy. +1
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      1. USR. Martinho Fernandes
    3. COThanks, answer to the point. I was confused as to whether this scenario of the looping variable incrementing as well decrementing while traversing throught the input can be termed as O(n). Your explanation helped. Thanks :-)
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